[seqfan] Re: reply to Message 4. Anyone recognize this matrix? (Neil Sloane)
I.V. Serov
i.v.serov at chf.nu
Thu Apr 22 22:25:53 CEST 2021
Dear Neil,
Would you allow to rearrange the order of the rows in the matrix?
It looks like the same rows can be generated (in a different order) by
means of a formula:
B(1) = 01.
B(2n+0) = concatenate(B'(n);B(n)).
B(2n+1) = concatenate(B(n);B(n)').
Here B(n)' equals B(n) with the rightmost bit flipped.
These are the first 31 rows of your matrix in the new order:
01*
0001*
0100*
00000001*
00010000*
01010100*
01000101*
0000000000000001*
0000000100000000*
0001000100010000*
0001000000010001*
0101010101010100*
0101010001010101*
0100010001000101*
0100010101000100*
00000000000000000000000000000001*
00000000000000010000000000000000*
00000001000000010000000100000000*
00000001000000000000000100000001*
00010001000100010001000100010000*
00010001000100000001000100010001*
00010000000100000001000000010001*
00010000000100010001000000010000*
01010101010101010101010101010100*
01010101010101000101010101010101*
01010100010101000101010001010101*
01010100010101010101010001010100*
01000100010001000100010001000101*
01000100010001010100010001000100*
01000101010001010100010101000100*
01000101010001000100010101000101*
The order permutations are described by A341915 and A341916.
Kind regards,
Igor Serov
www.chf.nu
On 21-04-2021 15:07, seqfan-request at list.seqfan.eu wrote:
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> Today's Topics:
>
> 1. A339950, A189378 (Don Reble)
> 2. Re: A339950, A189378 (Jeffrey Shallit)
> 3. How to define analogue to this sequence pertaining to Roman
> numerals (Alonso Del Arte)
> 4. Anyone recognize this matrix? (Neil Sloane)
> 5. Recent perturbations on the Seqfan Mailing List (Olivier Gerard)
> 6. Conjecture: a(n) = n only if n = 1 or 9 (Alonso Del Arte)
> 7. Lucasian (pseudo)primes (Tomasz Ordowski)
> 8. Re: Conjecture: a(n) = n only if n = 1 or 9 (D. S. McNeil)
> 9. Sum-Product Problem (Frank Adams-watters)
> 10. Planar distributive lattices (Allan Wechsler)
> 11. Re: Sum-Product Problem (Neil Sloane)
> 12. Re: Sum-Product Problem (Neil Sloane)
> 13. Re: Planar distributive lattices (Neil Sloane)
> 14. Re: Planar distributive lattices (Neil Sloane)
> 15. Re: Sum-Product Problem (Hugo Pfoertner)
> 16. Re: Conjecture: a(n) = n only if n = 1 or 9 (Alonso Del Arte)
> 17. Re: Planar distributive lattices (Allan Wechsler)
> 18. Re: Planar distributive lattices (Neil Sloane)
> 19. help naming/describing sequences for bounds of Goldbach's
> Comet, was: what happened to my A342302, its been replaced ???
> (peter lawrence)
> 20. Re: Anyone recognize this matrix? (Richard J. Mathar)
> 21. Re: Anyone recognize this matrix? (Neil Sloane)
>
>
> ----------------------------------------------------------------------
>
> Message: 1
> Date: Tue, 6 Apr 2021 02:19:20 -0600
> From: Don Reble <djr at nk.ca>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Subject: [seqfan] A339950, A189378
> Message-ID: <606C1988.90801 at nk.ca>
> Content-Type: text/plain; charset=UTF-8; format=flowed
>
> Seqfans:
>
> I computed more of A339950 (1,7,14,20,...,391,397,404).
> So far, A339950(n+1) = A189378(n)+1.
> Does anyone see how to (dis)prove that?
>
> --
> Don Reble djr at nk.ca
>
>
>
> ------------------------------
>
> Message: 2
> Date: Tue, 6 Apr 2021 13:10:45 -0400
> From: Jeffrey Shallit <shallit at uwaterloo.ca>
> To: <seqfan at list.seqfan.eu>
> Subject: [seqfan] Re: A339950, A189378
> Message-ID: <9ad00bb5-2f24-5cd5-3014-591013221ab3 at uwaterloo.ca>
> Content-Type: text/plain; charset="utf-8"; format=flowed
>
> Yes, I proved it this morning with the automatic theorem-prover
> "Walnut".
>
> On 2021-04-06 4:19 a.m., Don Reble via SeqFan wrote:
>> Seqfans:
>>
>> ? I computed more of A339950 (1,7,14,20,...,391,397,404).
>> ? So far, A339950(n+1) = A189378(n)+1.
>> ? Does anyone see how to (dis)prove that?
>>
>
>
> ------------------------------
>
> Message: 3
> Date: Thu, 8 Apr 2021 17:12:24 -0400
> From: Alonso Del Arte <alonso.delarte at gmail.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Subject: [seqfan] How to define analogue to this sequence pertaining
> to Roman numerals
> Message-ID:
> <CAGyGvfVOmJxnkd6s3Zuq5pNvOHDYVV2yeWrNPkYQQgHYBNN28A at mail.gmail.com>
> Content-Type: text/plain; charset="UTF-8"
>
> Almost all integers from 1 to 3999 are Roman numeral Harshad numbers.
> Much
> fewer of them can have their Roman numeral representations built up
> from
> the Roman numeral representations of their nontrivial divisors. For
> example, the divisors of 80 include 20 and 40, which in Roman numerals
> are
> XX and XL, respectively. From those we can assemble LXXX. I'm not
> interested in requiring that all the "digits" of a divisor be used,
> e.g.,
> for LXXX we could use X, XX and XL, using XL only for the L and
> discarding
> the extra X.
>
> scala> (1 to 3999).filter(romNumDivAnagrammable)
> res12: IndexedSeq[Int] = Vector(8, 18, 36, 80, 84, 88, 180, 184, 186,
> 270,
> 276, 282, 288, 360, 372, 380, 384, 396, 800, 804, 808, 810, 812, 816,
> 820,
> 822, 828, 832, 834, 836, 840, 846, 848, 852, 858, 860, 864, 868, 870,
> 876,
> 880, 882, 884, 888, 894, 896, 1800, 1804, 1806, 1808, 1812, 1816, 1818,
> 1820, 1824, 1830, 1832, 1836, 1840, 1848, 1856, 1860, 1864, 1872, 1876,
> 1880, 1884, 1888, 1890, 1896, 2700, 2706, 2712, 2718, 2724, 2730, 2736,
> 2742, 2748, 2754, 2760, 2766, 2772, 2778, 2784, 2790, 2796, 2802, 2808,
> 2814, 2820, 2826, 2832, 2838, 2844, 2850, 2856, 2862, 2868, 2874, 2880,
> 2886, 2892, 2898, 3600, 3612, 3624, 3636, 3648, 3660, 3672, 3684, 3696,
> 3708, 3720, 3732, 3744, 3756, 3768, 3780, 3792, 3800, 3804, 3808, 3810,
> 3816, 3820, 3822, 3824, 3828, 3832, 3834, 3836, 3840, 3846, 3848, 38...
>
> scala> romNums(res21)
> res23: IndexedSeq[numerics.RomanNumeralsNumber] = Vector(MMMDCCCXXIV,
> MMMDCCCXXVIII, MMMDCCCXXXII, MMMDCCCXXXIV, MMMDCCCXXXVI, MMMDCCCXL,
> MMMDCCCXLVI, MMMDCCCXLVIII, MMMDCCCLII, MMMDCCCLVI, MMMDCCCLVIII,
> MMMDCCCLX, MMMDCCCLXIV, MMMDCCCLXX, MMMDCCCLXXII, MMMDCCCLXXVI,
> MMMDCCCLXXX, MMMDCCCLXXXII, MMMDCCCLXXXVIII, MMMDCCCXCII, MMMDCCCXCIV,
> MMMCM, MMMCMXII, MMMCMXXIV, MMMCMXXXVI, MMMCMXLVIII, MMMCMLX,
> MMMCMLXXII,
> MMMCMLXXXIV, MMMCMXCVI)
>
> scala> romNums(res18)
> res24: IndexedSeq[numerics.RomanNumeralsNumber] = Vector(VIII, XVIII,
> XXXVI, LXXX, LXXXIV, LXXXVIII, CLXXX, CLXXXIV, CLXXXVI, CCLXX, CCLXXVI,
> CCLXXXII, CCLXXXVIII, CCCLX, CCCLXXII, CCCLXXX, CCCLXXXIV, CCCXCVI,
> DCCC,
> DCCCIV, DCCCVIII, DCCCX, DCCCXII, DCCCXVI, DCCCXX, DCCCXXII,
> DCCCXXVIII,
> DCCCXXXII, DCCCXXXIV, DCCCXXXVI, DCCCXL, DCCCXLVI, DCCCXLVIII, DCCCLII,
> DCCCLVIII, DCCCLX, DCCCLXIV, DCCCLXVIII, DCCCLXX, DCCCLXXVI, DCCCLXXX,
> DCCCLXXXII, DCCCLXXXIV, DCCCLXXXVIII, DCCCXCIV, DCCCXCVI, MDCCC,
> MDCCCIV,
> MDCCCVI, MDCCCVIII, MDCCCXII, MDCCCXVI, MDCCCXVIII, MDCCCXX, MDCCCXXIV,
> MDCCCXXX, MDCCCXXXII, MDCCCXXXVI, MDCCCXL, MDCCCXLVIII, MDCCCLVI,
> MDCCCLX,
> MDCCCLXIV, MDCCCLXXII, MDCCCLXXVI, MDCCCLXXX, MDCCCLXXXIV,
> MDCCCLXXXVIII,
> MDCCCXC, MDCCCXCVI, MMDCC, MMDCCVI, MMDCCXII, MMDCCXVIII, MMDCCXXIV,
> M...
>
> I wrote the Boolean romNumDivAnagrammable() function to only consider
> the
> divisors of *n* other than 1 and *n* itself. If I haven't made a
> mistake
> somewhere, I can assert that this doesn't make a difference for
> "divisor
> anagrammables" in Roman numerals. Taking this to decimal, we see that
> 12
> would not thus be "divisor anagrammable," since the divisor 1 is not
> considered. Thoughts?
>
> Al
>
> --
> Alonso del Arte
> Author at SmashWords.com
> <https://www.smashwords.com/profile/view/AlonsoDelarte>
> Musician at ReverbNation.com
> <http://www.reverbnation.com/alonsodelarte>
>
>
> ------------------------------
>
> Message: 4
> Date: Fri, 9 Apr 2021 22:06:31 -0400
> From: Neil Sloane <njasloane at gmail.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Subject: [seqfan] Anyone recognize this matrix?
> Message-ID:
> <CAAOnSgS89HNuj2uW5NMGx_qCPV8mu_h_Gi=wLJR-eROGQXhzVA at mail.gmail.com>
> Content-Type: text/plain; charset="UTF-8"
>
> Dear Sequence Fans, I have an infinite 0,1 matrix. The first row is 01
> repeated, the second row is 0100 repeated, and so on. Here are the
> first 32
> rows.
> I have a feeling I've seen this before, but I can't remember where.
> I have the definition, but I would like a simple description.
> Does anyone recognize this?
>
> There are some obvious properties. In rows 8 through 15, for instance,
> the
> mod 2 sums row 8 + row 15 = row 9 + row 14 = ... = row 11 + row 12 =
> 0000000100000001.
> And similarly for rows 2 to 3; 4 to 7; 16 to 31; etc.
>
>
> 1: 01*
> 2: 0100*
> 3: 0001*
> 4: 00010000*
> 5: 01000101*
> 6: 01010100*
> 7: 00000001*
> 8: 0000000100000000*
> 9: 0101010001010101*
> 10: 0100010101000100*
> 11: 0001000000010001*
> 12: 0001000100010000*
> 13: 0100010001000101*
> 14: 0101010101010100*
> 15: 0000000000000001*
> 16: 00000000000000010000000000000000*
> 17: 01010101010101000101010101010101*
> 18: 01000100010001010100010001000100*
> 19: 00010001000100000001000100010001*
> 20: 00010000000100010001000000010000*
> 21: 01000101010001000100010101000101*
> 22: 01010100010101010101010001010100*
> 23: 00000001000000000000000100000001*
> 24: 00000001000000010000000100000000*
> 25: 01010100010101000101010001010101*
> 26: 01000101010001010100010101000100*
> 27: 00010000000100000001000000010001*
> 28: 00010001000100010001000100010000*
> 29: 01000100010001000100010001000101*
> 30: 01010101010101010101010101010100*
> 31: 00000000000000000000000000000001*
>
> [These are actually the odd-numbered rows 1,3,5,7,... of the matrix.
> The
> even-numbered rows have a simple formula. Row 2k is 0^(2^m) 1^(2^m)
> repeated, where m is the number of times 2 divides 2k.
>
> Row 24 for example (where m=3) is 0000000011111111 repeated. I'm hoping
> for
> something similar for the odd-numbered rows.]
>
>
> ------------------------------
>
> Message: 5
> Date: Sat, 17 Apr 2021 17:55:15 +0300
> From: Olivier Gerard <olivier.gerard at gmail.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Subject: [seqfan] Recent perturbations on the Seqfan Mailing List
> Message-ID:
> <CAAcpPSYWHXp6ZMaMVd4-K0-_EVACNRdXgPtyAhq79Z5ywjRhrA at mail.gmail.com>
> Content-Type: text/plain; charset="UTF-8"
>
> Dear subscribers of the Seqfan Mailing List,
>
> This is a test message as our server has suffered from technical
> glitches
> lately
> and they may reappear and take some time to calm down.
>
> Please do not respond to this message on the list, but you can send a
> private message
> to me if you want to.
>
> With all my apologies for any inconvenience it may have caused.
>
> With my best regards,
>
> Olivier GERARD
> Seqfan Mailing List Administrator
> olivier.gerard at gmail.com
>
>
> ------------------------------
>
> Message: 6
> Date: Thu, 15 Apr 2021 16:56:06 -0400
> From: Alonso Del Arte <alonso.delarte at gmail.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Subject: [seqfan] Conjecture: a(n) = n only if n = 1 or 9
> Message-ID:
> <CAGyGvfUowAjOVEE8q1Ac6AhH5BNJv0wUREBKBwxV=b_ypkBBoQ at mail.gmail.com>
> Content-Type: text/plain; charset="UTF-8"
>
> Given a(n) = a(n) = \prod_{i = 1}^{n - 1} gcd(i, n) (A051190) it is
> obvious
> that a(n) = 1 only if n is 1 or a prime.
>
> For many OEIS entries Charles has given a heuristic for growth, though
> not
> for this one. It seems to be a decent fraction of n! for composite n.
>
> Of course this doesn't rule out that there might be some larger n such
> that
> a(n) = n.
>
> Thoughts, anyone?
>
> Al
>
> --
> Alonso del Arte
> Author at SmashWords.com
> <https://www.smashwords.com/profile/view/AlonsoDelarte>
> Musician at ReverbNation.com
> <http://www.reverbnation.com/alonsodelarte>
>
>
> ------------------------------
>
> Message: 7
> Date: Sat, 17 Apr 2021 17:45:25 +0200
> From: Tomasz Ordowski <tomaszordowski at gmail.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Subject: [seqfan] Lucasian (pseudo)primes
> Message-ID:
> <CAF0qcNN0Q690OHqXH=FQvidTXa+hnfEYubK6fOHTDfNE1cETOA at mail.gmail.com>
> Content-Type: text/plain; charset="UTF-8"
>
> Dear readers!
>
> If p is a Lucasian prime, i.e.
> p == 3 (mod 4) with 2p+1 prime;
> then (2^p-1)/(2p+1) == 1 (mod p),
> hence 2^p-2p-2 == 0 (mod p(2p+1)),
> so 2^(p-1) == p+1 (mod p(2p+1)).
>
> Composites k such that 2^(k-1) == k+1 (mod k(2k+1)) are
> 150851, 452051, 1325843, 1441091, 4974971, 5016191, 15139199, 19020191,
> 44695211, 101276579, 119378351, 128665319, 152814531, 187155383,
> 203789951,
> 223782263, 307367171, 387833531, 392534231, 470579831, 505473263,
> 546748931, 626717471, 639969891, 885510239, 974471243, 1147357559,
> 1227474431, 1284321611, 1304553251, 1465307351, 1474936871, 1514608559,
> 1529648231, 1639846391, 1672125131, 2117031263, 2139155051, 2304710123,
> 2324867399, 2939179643, 3056100623, 3271076771, 3280593611, 3529864391,
> 3587553971, 4193496803, 4244663651, 4267277291, 4278305651, 4528686251,
> ...
> [Data from Amiram Eldar]
>
> Conjecture:
> These are pseudoprimes k == 3 (mod 4) such that 2k+1 is prime.
> If so, the name "Lucasian pseudoprimes" will be fully justified.
>
> There are 101629 Fermat pseudoprimes up to 10^12.
> Of them 276 are of the type k == 3 (mod 4) with 2k+1 prime.
> The first 51 of them are exactly those that I have sent you earlier.
> [Amiram Eldar]
>
> The conjecture is easily proved.
> Let q = 2k+1 be prime, where k == 3 (mod 4) is a pseudoprime.
> We have q == 7 (mod 8), so 2 is a square mod q,
> which gives 2^{(q-1)/2} == 1 (mod q), by Euler's criterion.
> Thus, 2^k == 1 (mod q), which implies 2^{k-1} == (q+1)/2 (mod q),
> so that 2^{k-1} == k+1 (mod q).
> The conclusion that 2^{k-1} == k+1 (mod kq) follows
> from the assumption that k is a pseudoprime
> and from the Chinese remainder theorem.
> [Carl Pomerance]
>
> Problem:
> Are there infinitely many numbers n such that 2^{n-1} == n+1 (mod
> n(2n+1))
> ?
> These are primes and pseudoprimes n == 3 (mod 4) with 2n+1 prime.
> It is not known whether there are infinitely many Lucasian primes.
>
> Question:
> Are there pseudoprimes m == 3 (mod 4) such that 2m+1 is a pseudoprime?
>
> There only 3 known pseudoprimes m such that 2m+1 is a pseudoprime:
> 9890881, 23456248059221, 96076792050570581 (see A303447),
> but all the three have m == 1 (mod 4).
> [Amiram Eldar]
>
> Best regards,
>
> Thomas Ordowski
> ____________________
> https://oeis.org/A002515
> https://oeis.org/A081858
> https://oeis.org/A001567
> https://oeis.org/A303447
>
>
> ------------------------------
>
> Message: 8
> Date: Sat, 17 Apr 2021 14:50:57 -0400
> From: "D. S. McNeil" <dsm054 at gmail.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Subject: [seqfan] Re: Conjecture: a(n) = n only if n = 1 or 9
> Message-ID:
> <CAOX9QiAL2+z-L_Fb9BPRHayH4Khncr2wfC-hmOyhKR-x8u4K5Q at mail.gmail.com>
> Content-Type: text/plain; charset="UTF-8"
>
> I think it's true. Sketch I'm too lazy to formalize (and whenever I
> break
> things apart by cases I feel like I'm missing something obvious):
>
> Say some composite n is a pure prime power p^k.
>
> For k=2, we pick up a factor of p every p in the product (except for
> p^2
> itself), and so A(p^2) = p^(p-1). The only solution p=3 generates the
> n=9
> solution. For p > 3, A(p^2) > n.
>
> For k=3, we pick up additional factors of p^2, and if I'm counting
> correctly A(p^3) = p^(p^2+p-2), which generates no solutions.
> Moreover,
> A(p^3) > n for all p.
>
> For k>=4, we have at least prod(p^i,i=1..k-1) as part of the product,
> and
> so p^((k-1)*k/2) | A(n). Since (k-1)*k/2 > k for k >= 4, this suffices
> to
> show A(n) > n.
>
> Now assume it's not a prime power, and we have n = p_0^k_0 p_1^k_1..
> p_m^k_m. It suffices to show A(p^k*q) > p^k*q for gcd(p, q) = 1.
>
> We'll pick up factors of p^k and q immediately in the expanded product,
> and
> thus p^k * q | A(p^k*q), so all we need is one other component to
> exceed
> p^k * q.
>
> If k>=2, p itself contributes in the terms: [p, p^k, q] giving p^(k+1)
> * q
> | A(n).
> If k = 1 and p=2, then q >= 3, n >= 6, and thus [p=2, 2*p=4, q] giving
> p^2*q | A(n).
> If k = 1 and q=2, then p >= 3, n >= 6, and thus [p, q=2, 2*q=4] giving
> p*q^2 | A(n)
> If k = 1 and p,q >= 3, then we have [p, 2*p, q] giving p^2 * q| A(n)
>
> And I think that's all the cases.
>
> If I didn't make a silly error in the above it'd be the first time
> ever,
> but I'm pretty sure some argument along these lines works.
>
>
> Doug #lockdownboredom
>
>
> ------------------------------
>
> Message: 9
> Date: Sat, 17 Apr 2021 17:37:37 +0000 (UTC)
> From: Frank Adams-watters <franktaw at netscape.net>
> To: "seqfan at seqfan.eu" <seqfan at seqfan.eu>
> Subject: [seqfan] Sum-Product Problem
> Message-ID: <499880394.1809449.1618681057196 at mail.yahoo.com>
> Content-Type: text/plain; charset=UTF-8
>
> There's an article on this problem in Quanta magazine:
>
> https://www.quantamagazine.org/the-sum-product-problem-shows-how-addition-and-multiplication-constrain-each-other-20190206/
>
> Should this be added as a link to A027424? Or do the academic links
> already there suffice?
>
> Franklin T. Adams-Watters
>
>
> ------------------------------
>
> Message: 10
> Date: Wed, 14 Apr 2021 21:08:03 -0400
> From: Allan Wechsler <acwacw at gmail.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Subject: [seqfan] Planar distributive lattices
> Message-ID:
> <CADy-sGHf=VWa90-vsBdqRTOCPFCNVL1B6ECknaOb7Xgr7Tt+9w at mail.gmail.com>
> Content-Type: text/plain; charset="UTF-8"
>
> I forget how I stumbled on this:
> https://math.chapman.edu/~jipsen/mathposters/Planar%20distributive%20lattices%20up%20to%20size%2011.pdf
> .
>
> It is a chart purporting to show all of the planar distributive
> lattices
> with up to 11 vertices. Like any true-hearted sequence fanatic I
> counted
> the number of these guys of each order, and got the following sequence:
>
> 1,1,1,2,3,5,8,14,24,42,72...
>
> Imagine my surprise at finding this sequence missing from OEIS! The
> author
> is apparently Dr. Peter Jipsen, at Chapman University in California.
>
> Perhaps someone here can figure out (a) what a planar distributive
> lattice
> is, (b) whether Dr. Jipsen enumerated them correctly, (c) whether I
> counted
> them off Jipsen's poster correctly, and (d) whether to add the
> sequence.
>
> Thank you!
>
>
> ------------------------------
>
> Message: 11
> Date: Sat, 17 Apr 2021 21:23:05 -0400
> From: Neil Sloane <njasloane at gmail.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Cc: "seqfan at seqfan.eu" <seqfan at seqfan.eu>, Frank Adams-watters
> <franktaw at netscape.net>
> Subject: [seqfan] Re: Sum-Product Problem
> Message-ID:
> <CAAOnSgSHELbp3auLvoS-DPK6bYbOnqTPWX8xj=1CJneBXa03eg at mail.gmail.com>
> Content-Type: text/plain; charset="UTF-8"
>
> certainly add it - thanks!
>
>
> Best regards
> Neil
>
> Neil J. A. Sloane, President, OEIS Foundation.
> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway,
> NJ.
> Phone: 732 828 6098; home page: http://NeilSloane.com
> Email: njasloane at gmail.com
>
>
>
> On Sat, Apr 17, 2021 at 9:21 PM Frank Adams-watters via SeqFan <
> seqfan at list.seqfan.eu> wrote:
>
>> There's an article on this problem in Quanta magazine:
>>
>>
>> https://www.quantamagazine.org/the-sum-product-problem-shows-how-addition-and-multiplication-constrain-each-other-20190206/
>>
>> Should this be added as a link to A027424? Or do the academic links
>> already there suffice?
>>
>> Franklin T. Adams-Watters
>>
>> --
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>
>
> ------------------------------
>
> Message: 12
> Date: Sat, 17 Apr 2021 21:23:05 -0400
> From: Neil Sloane <njasloane at gmail.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Cc: "seqfan at seqfan.eu" <seqfan at seqfan.eu>, Frank Adams-watters
> <franktaw at netscape.net>
> Subject: [seqfan] Re: Sum-Product Problem
> Message-ID:
> <CAAOnSgSHELbp3auLvoS-DPK6bYbOnqTPWX8xj=1CJneBXa03eg at mail.gmail.com>
> Content-Type: text/plain; charset="UTF-8"
>
> certainly add it - thanks!
>
>
> Best regards
> Neil
>
> Neil J. A. Sloane, President, OEIS Foundation.
> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway,
> NJ.
> Phone: 732 828 6098; home page: http://NeilSloane.com
> Email: njasloane at gmail.com
>
>
>
> On Sat, Apr 17, 2021 at 9:21 PM Frank Adams-watters via SeqFan <
> seqfan at list.seqfan.eu> wrote:
>
>> There's an article on this problem in Quanta magazine:
>>
>>
>> https://www.quantamagazine.org/the-sum-product-problem-shows-how-addition-and-multiplication-constrain-each-other-20190206/
>>
>> Should this be added as a link to A027424? Or do the academic links
>> already there suffice?
>>
>> Franklin T. Adams-Watters
>>
>> --
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>
>
> ------------------------------
>
> Message: 13
> Date: Sun, 18 Apr 2021 01:17:36 -0400
> From: Neil Sloane <njasloane at gmail.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Subject: [seqfan] Re: Planar distributive lattices
> Message-ID:
> <CAAOnSgQDtfv9ZpBauaYJqPBveVKc=98pZ00rY_uKZ3f=g=_MwA at mail.gmail.com>
> Content-Type: text/plain; charset="UTF-8"
>
> Allan, I confirm your numbers. I'll run it through Superseeker, which
> will
> tell us if it is a simple cousin of an existing entry, and if it
> doesn;t
> find anything I will add it - it will be A343161.
>
> Thanks for catching this fish!
>
> Best regards
> Neil
>
> Neil J. A. Sloane, President, OEIS Foundation.
> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway,
> NJ.
> Phone: 732 828 6098; home page: http://NeilSloane.com
> Email: njasloane at gmail.com
>
>
>
> On Sat, Apr 17, 2021 at 9:22 PM Allan Wechsler <acwacw at gmail.com>
> wrote:
>
>> I forget how I stumbled on this:
>>
>> https://math.chapman.edu/~jipsen/mathposters/Planar%20distributive%20lattices%20up%20to%20size%2011.pdf
>> .
>>
>> It is a chart purporting to show all of the planar distributive
>> lattices
>> with up to 11 vertices. Like any true-hearted sequence fanatic I
>> counted
>> the number of these guys of each order, and got the following
>> sequence:
>>
>> 1,1,1,2,3,5,8,14,24,42,72...
>>
>> Imagine my surprise at finding this sequence missing from OEIS! The
>> author
>> is apparently Dr. Peter Jipsen, at Chapman University in California.
>>
>> Perhaps someone here can figure out (a) what a planar distributive
>> lattice
>> is, (b) whether Dr. Jipsen enumerated them correctly, (c) whether I
>> counted
>> them off Jipsen's poster correctly, and (d) whether to add the
>> sequence.
>>
>> Thank you!
>>
>> --
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>
>
> ------------------------------
>
> Message: 14
> Date: Sun, 18 Apr 2021 01:38:57 -0400
> From: Neil Sloane <njasloane at gmail.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Subject: [seqfan] Re: Planar distributive lattices
> Message-ID:
> <CAAOnSgR_QP5B7hMVB7KZ+=tdP4eSPqFoYnhVd7+6PfiDeXvT=Q at mail.gmail.com>
> Content-Type: text/plain; charset="UTF-8"
>
> PS I see that the author of that poster updated it in 2014, extending
> it to
> 15 vertices:
>
> https://math.chapman.edu/~jipsen/tikzsvg/planar-distributive-lattices15.html
>
> I am going to write to him
>
>
> Best regards
> Neil
>
> Neil J. A. Sloane, President, OEIS Foundation.
> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway,
> NJ.
> Phone: 732 828 6098; home page: http://NeilSloane.com
> Email: njasloane at gmail.com
>
>
>
> On Sun, Apr 18, 2021 at 1:17 AM Neil Sloane <njasloane at gmail.com>
> wrote:
>
>> Allan, I confirm your numbers. I'll run it through Superseeker, which
>> will tell us if it is a simple cousin of an existing entry, and if it
>> doesn;t find anything I will add it - it will be A343161.
>>
>> Thanks for catching this fish!
>>
>> Best regards
>> Neil
>>
>> Neil J. A. Sloane, President, OEIS Foundation.
>> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
>> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway,
>> NJ.
>> Phone: 732 828 6098; home page: http://NeilSloane.com
>> Email: njasloane at gmail.com
>>
>>
>>
>> On Sat, Apr 17, 2021 at 9:22 PM Allan Wechsler <acwacw at gmail.com>
>> wrote:
>>
>>> I forget how I stumbled on this:
>>>
>>> https://math.chapman.edu/~jipsen/mathposters/Planar%20distributive%20lattices%20up%20to%20size%2011.pdf
>>> .
>>>
>>> It is a chart purporting to show all of the planar distributive
>>> lattices
>>> with up to 11 vertices. Like any true-hearted sequence fanatic I
>>> counted
>>> the number of these guys of each order, and got the following
>>> sequence:
>>>
>>> 1,1,1,2,3,5,8,14,24,42,72...
>>>
>>> Imagine my surprise at finding this sequence missing from OEIS! The
>>> author
>>> is apparently Dr. Peter Jipsen, at Chapman University in California.
>>>
>>> Perhaps someone here can figure out (a) what a planar distributive
>>> lattice
>>> is, (b) whether Dr. Jipsen enumerated them correctly, (c) whether I
>>> counted
>>> them off Jipsen's poster correctly, and (d) whether to add the
>>> sequence.
>>>
>>> Thank you!
>>>
>>> --
>>> Seqfan Mailing list - http://list.seqfan.eu/
>>>
>>
>
>
> ------------------------------
>
> Message: 15
> Date: Sun, 18 Apr 2021 11:10:42 +0200
> From: Hugo Pfoertner <yae9911 at gmail.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Subject: [seqfan] Re: Sum-Product Problem
> Message-ID:
> <CAEoHttX4cE=oND4TxkevV-6knbq1=QVntzDBiBQfd2afNELtnQ at mail.gmail.com>
> Content-Type: text/plain; charset="UTF-8"
>
> A263996 <https://oeis.org/A263996> is even more closely related. I also
> added the link there.
>
> On Sun, Apr 18, 2021 at 3:23 AM Neil Sloane <njasloane at gmail.com>
> wrote:
>
>> certainly add it - thanks!
>>
>>
>> Best regards
>> Neil
>>
>> Neil J. A. Sloane, President, OEIS Foundation.
>> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
>> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway,
>> NJ.
>> Phone: 732 828 6098; home page: http://NeilSloane.com
>> Email: njasloane at gmail.com
>>
>>
>>
>> On Sat, Apr 17, 2021 at 9:21 PM Frank Adams-watters via SeqFan <
>> seqfan at list.seqfan.eu> wrote:
>>
>> > There's an article on this problem in Quanta magazine:
>> >
>> >
>> >
>> https://www.quantamagazine.org/the-sum-product-problem-shows-how-addition-and-multiplication-constrain-each-other-20190206/
>> >
>> > Should this be added as a link to A027424? Or do the academic links
>> > already there suffice?
>> >
>> > Franklin T. Adams-Watters
>> >
>> > --
>> > Seqfan Mailing list - http://list.seqfan.eu/
>> >
>>
>> --
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>
>
> ------------------------------
>
> Message: 16
> Date: Sun, 18 Apr 2021 14:05:21 -0400
> From: Alonso Del Arte <alonso.delarte at gmail.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Subject: [seqfan] Re: Conjecture: a(n) = n only if n = 1 or 9
> Message-ID:
> <CAGyGvfXhq4Ae4xRXfF=mK5UGjs4aMTV7zDqOek=yrnQ5Jk0xVg at mail.gmail.com>
> Content-Type: text/plain; charset="UTF-8"
>
> Thanks, D. S., I think it checks out. I appreciate your taking the time
> to
> sketch this out. In all honesty, the most I did was scan the B-file.
>
> Al
>
> On Sat, Apr 17, 2021 at 2:51 PM D. S. McNeil <dsm054 at gmail.com> wrote:
>
>> I think it's true. Sketch I'm too lazy to formalize (and whenever I
>> break
>> things apart by cases I feel like I'm missing something obvious):
>>
>> Say some composite n is a pure prime power p^k.
>>
>> For k=2, we pick up a factor of p every p in the product (except for
>> p^2
>> itself), and so A(p^2) = p^(p-1). The only solution p=3 generates the
>> n=9
>> solution. For p > 3, A(p^2) > n.
>>
>> For k=3, we pick up additional factors of p^2, and if I'm counting
>> correctly A(p^3) = p^(p^2+p-2), which generates no solutions.
>> Moreover,
>> A(p^3) > n for all p.
>>
>> For k>=4, we have at least prod(p^i,i=1..k-1) as part of the product,
>> and
>> so p^((k-1)*k/2) | A(n). Since (k-1)*k/2 > k for k >= 4, this
>> suffices to
>> show A(n) > n.
>>
>> Now assume it's not a prime power, and we have n = p_0^k_0 p_1^k_1..
>> p_m^k_m. It suffices to show A(p^k*q) > p^k*q for gcd(p, q) = 1.
>>
>> We'll pick up factors of p^k and q immediately in the expanded
>> product, and
>> thus p^k * q | A(p^k*q), so all we need is one other component to
>> exceed
>> p^k * q.
>>
>> If k>=2, p itself contributes in the terms: [p, p^k, q] giving p^(k+1)
>> * q
>> | A(n).
>> If k = 1 and p=2, then q >= 3, n >= 6, and thus [p=2, 2*p=4, q] giving
>> p^2*q | A(n).
>> If k = 1 and q=2, then p >= 3, n >= 6, and thus [p, q=2, 2*q=4] giving
>> p*q^2 | A(n)
>> If k = 1 and p,q >= 3, then we have [p, 2*p, q] giving p^2 * q| A(n)
>>
>> And I think that's all the cases.
>>
>> If I didn't make a silly error in the above it'd be the first time
>> ever,
>> but I'm pretty sure some argument along these lines works.
>>
>>
>> Doug #lockdownboredom
>>
>> --
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>
>
> --
> Alonso del Arte
> Author at SmashWords.com
> <https://www.smashwords.com/profile/view/AlonsoDelarte>
> Musician at ReverbNation.com
> <http://www.reverbnation.com/alonsodelarte>
>
>
> ------------------------------
>
> Message: 17
> Date: Sun, 18 Apr 2021 17:36:45 -0400
> From: Allan Wechsler <acwacw at gmail.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Subject: [seqfan] Re: Planar distributive lattices
> Message-ID:
> <CADy-sGG_avdNiEmVMj2=8NM34CiVRO_O=SRWWaE9suFNJwzo=A at mail.gmail.com>
> Content-Type: text/plain; charset="UTF-8"
>
> I think writing to him is a great idea; get him to author that
> sequence,
> which he only deserves, after all.
>
> On Sun, Apr 18, 2021 at 1:39 AM Neil Sloane <njasloane at gmail.com>
> wrote:
>
>> PS I see that the author of that poster updated it in 2014, extending
>> it to
>> 15 vertices:
>>
>>
>> https://math.chapman.edu/~jipsen/tikzsvg/planar-distributive-lattices15.html
>>
>> I am going to write to him
>>
>>
>> Best regards
>> Neil
>>
>> Neil J. A. Sloane, President, OEIS Foundation.
>> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
>> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway,
>> NJ.
>> Phone: 732 828 6098; home page: http://NeilSloane.com
>> Email: njasloane at gmail.com
>>
>>
>>
>> On Sun, Apr 18, 2021 at 1:17 AM Neil Sloane <njasloane at gmail.com>
>> wrote:
>>
>> > Allan, I confirm your numbers. I'll run it through Superseeker, which
>> > will tell us if it is a simple cousin of an existing entry, and if it
>> > doesn;t find anything I will add it - it will be A343161.
>> >
>> > Thanks for catching this fish!
>> >
>> > Best regards
>> > Neil
>> >
>> > Neil J. A. Sloane, President, OEIS Foundation.
>> > 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
>> > Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
>> > Phone: 732 828 6098; home page: http://NeilSloane.com
>> > Email: njasloane at gmail.com
>> >
>> >
>> >
>> > On Sat, Apr 17, 2021 at 9:22 PM Allan Wechsler <acwacw at gmail.com> wrote:
>> >
>> >> I forget how I stumbled on this:
>> >>
>> >>
>> https://math.chapman.edu/~jipsen/mathposters/Planar%20distributive%20lattices%20up%20to%20size%2011.pdf
>> >> .
>> >>
>> >> It is a chart purporting to show all of the planar distributive lattices
>> >> with up to 11 vertices. Like any true-hearted sequence fanatic I counted
>> >> the number of these guys of each order, and got the following sequence:
>> >>
>> >> 1,1,1,2,3,5,8,14,24,42,72...
>> >>
>> >> Imagine my surprise at finding this sequence missing from OEIS! The
>> author
>> >> is apparently Dr. Peter Jipsen, at Chapman University in California.
>> >>
>> >> Perhaps someone here can figure out (a) what a planar distributive
>> lattice
>> >> is, (b) whether Dr. Jipsen enumerated them correctly, (c) whether I
>> >> counted
>> >> them off Jipsen's poster correctly, and (d) whether to add the sequence.
>> >>
>> >> Thank you!
>> >>
>> >> --
>> >> Seqfan Mailing list - http://list.seqfan.eu/
>> >>
>> >
>>
>> --
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>
>
> ------------------------------
>
> Message: 18
> Date: Sun, 18 Apr 2021 22:10:06 -0400
> From: Neil Sloane <njasloane at gmail.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Subject: [seqfan] Re: Planar distributive lattices
> Message-ID:
> <CAAOnSgRb+u4ToEYxTM1+F2ns-7WHoQFr6vmsf=GnF8f0MhguLg at mail.gmail.com>
> Content-Type: text/plain; charset="UTF-8"
>
> Allan Wechsler said, in connection with the sequence I created
> (A343161)
> based on Peter Jipsen's enumeration: "I think writing to him is a
> great
> idea; get him to author that sequence, which he only deserves, after
> all."
> Well, he has been a registered user - and contributor - to the OEIS
> since
> March 2013. He never submitted it, and Allan wasn't sure it should be
> in
> the OEIS, so I created it. Today I extended it to 15 terms using the
> data
> in Peter's 2014 pdf file.
>
> Incidentally Peter thanked me for creating the entry and said in reply
> that
> he will try to find his old program and compute more terms.
>
> Best regards
> Neil
>
> Neil J. A. Sloane, President, OEIS Foundation.
> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway,
> NJ.
> Phone: 732 828 6098; home page: http://NeilSloane.com
> Email: njasloane at gmail.com
>
>
>
> On Sun, Apr 18, 2021 at 7:52 PM Allan Wechsler <acwacw at gmail.com>
> wrote:
>
>> I think writing to him is a great idea; get him to author that
>> sequence,
>> which he only deserves, after all.
>>
>> On Sun, Apr 18, 2021 at 1:39 AM Neil Sloane <njasloane at gmail.com>
>> wrote:
>>
>> > PS I see that the author of that poster updated it in 2014, extending it
>> to
>> > 15 vertices:
>> >
>> >
>> >
>> https://math.chapman.edu/~jipsen/tikzsvg/planar-distributive-lattices15.html
>> >
>> > I am going to write to him
>> >
>> >
>> > Best regards
>> > Neil
>> >
>> > Neil J. A. Sloane, President, OEIS Foundation.
>> > 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
>> > Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
>> > Phone: 732 828 6098; home page: http://NeilSloane.com
>> > Email: njasloane at gmail.com
>> >
>> >
>> >
>> > On Sun, Apr 18, 2021 at 1:17 AM Neil Sloane <njasloane at gmail.com> wrote:
>> >
>> > > Allan, I confirm your numbers. I'll run it through Superseeker, which
>> > > will tell us if it is a simple cousin of an existing entry, and if it
>> > > doesn;t find anything I will add it - it will be A343161.
>> > >
>> > > Thanks for catching this fish!
>> > >
>> > > Best regards
>> > > Neil
>> > >
>> > > Neil J. A. Sloane, President, OEIS Foundation.
>> > > 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
>> > > Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway,
>> NJ.
>> > > Phone: 732 828 6098; home page: http://NeilSloane.com
>> > > Email: njasloane at gmail.com
>> > >
>> > >
>> > >
>> > > On Sat, Apr 17, 2021 at 9:22 PM Allan Wechsler <acwacw at gmail.com>
>> wrote:
>> > >
>> > >> I forget how I stumbled on this:
>> > >>
>> > >>
>> >
>> https://math.chapman.edu/~jipsen/mathposters/Planar%20distributive%20lattices%20up%20to%20size%2011.pdf
>> > >> .
>> > >>
>> > >> It is a chart purporting to show all of the planar distributive
>> lattices
>> > >> with up to 11 vertices. Like any true-hearted sequence fanatic I
>> counted
>> > >> the number of these guys of each order, and got the following
>> sequence:
>> > >>
>> > >> 1,1,1,2,3,5,8,14,24,42,72...
>> > >>
>> > >> Imagine my surprise at finding this sequence missing from OEIS! The
>> > author
>> > >> is apparently Dr. Peter Jipsen, at Chapman University in California.
>> > >>
>> > >> Perhaps someone here can figure out (a) what a planar distributive
>> > lattice
>> > >> is, (b) whether Dr. Jipsen enumerated them correctly, (c) whether I
>> > >> counted
>> > >> them off Jipsen's poster correctly, and (d) whether to add the
>> sequence.
>> > >>
>> > >> Thank you!
>> > >>
>> > >> --
>> > >> Seqfan Mailing list - http://list.seqfan.eu/
>> > >>
>> > >
>> >
>> > --
>> > Seqfan Mailing list - http://list.seqfan.eu/
>> >
>>
>> --
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>
>
> ------------------------------
>
> Message: 19
> Date: Sun, 18 Apr 2021 21:54:25 -0700
> From: peter lawrence <peterl95124 at comcast.net>
> To: seqfan at list.seqfan.eu
> Cc: Olivier Gerard <olivier.gerard at gmail.com>, Susanna Cuyler
> <Susanna.Cuyler at gmail.com>, Neil Sloane <njasloane at gmail.com>
> Subject: [seqfan] help naming/describing sequences for bounds of
> Goldbach's Comet, was: what happened to my A342302, its been replaced
> ???
> Message-ID: <5B52ED1E-B02E-42B9-A71C-7E5AA36BCEE5 at comcast.net>
> Content-Type: text/plain; charset=utf-8
>
> All,
> I agree that the encyclopedia itself maybe isn?t the right place
> to work out the details,
> so lets do that here in seqfan instead.
>
> Goldbach's Comet (e.g.
> https://en.wikipedia.org/wiki/Goldbach%27s_comet
> <https://en.wikipedia.org/wiki/Goldbach's_comet>, and lots of google
> images, etc)
> has obvious upper and lower bounds, at least visually, so the natural
> question is what are those bounds.
>
> I?m using A002372 for Goldbach counts (order dependent sums) because
> they can be computed as a convolution,
> and because there?s something obvious about when the value is even
> verse odd that is much less easy to state
> when using A002375 (unordered sums).
>
> note that A002372 isn't indexed by N, rather it's indexed by N/2
> because odd numbers aren?t generally
> the sum of two primes (in general only the upper half of a twin prime
> pair is), and the factor of two
> makes it awkward to relate that sequence to ones discussed here where
> I talk about what N achieves a
> bound rather than what N/2.
>
> it would be nice to have approximation formulas,
> and it would also be nice to have sequences that could be found in
> OEIS,
> I?d like to work on both, but for this email I?ll stick with sequences.
>
> for the upper bound sequence one might tabulate where a high point is
> first achieved, as in
>
> 1 is first achieved at N = 6, 6 = {3+3} = 1 way
> 2 is first achieved at N = 8, 8 = {3+5, 5+3} = 2 ways
> 3 is first achieved at N = 10, 10 = {3+7, 5+5, 7+3} = 3 ways
> 4 is first achieved at N = 16, 16 = {3+13, 5+11, 11+5, 13+3} = 4 ways
> 5 is first achieved at N = 22, 22 = {3+19, 5+17, 11+11, 17+5, 19+3} =
> 5 ways
> etc...
>
> so 6,8,10,16,22,... maybe should be in OEIS ?
>
> well not so fast, continuing we find that the first and last N for
> each possible count 1..20 are
> count first last
> 1 6 6
> 2 8 12
> 3 10 38
> 4 16 68
> 5 22 62
> 6 24 128
> 7 34 122
> 8 36 152
> 9 74 158 --- 9 isn?t achieved until N=74, but a couple
> higher counts (10,120 are achieved earlier
> 10 48 188
> 11 106 166 --- ditto
> 12 60 332
> 13 178 398 --- and so it goes for all odd counts since an
> odd count only
> 14 78 272
> 15 142 362 --- occurs for N = 2 x Prime which are more rare
> than N = 2 x Composite
> 16 84 368
> 17 202 458
> 18 90 488
> 19 358 542
> 20 114 632
>
> so the strict upper bounds (points on the (top side of the) convex
> hull of Goldbach?s Comet)
> don?t include all possible counts, example ?9? above.
>
> also note that for just even counts the first occurrence isn?t always
> steadily increasing
> 30 234 908
> 32 246 1112
> 34 288 968
> 36 240 1412 ? the count 36 occurs earlier than 32, and 34
> 38 210 1178 ? the count 38 occurs earlier than 30, 32, 34, and
> 36
> 40 324 1448
>
> so not all even counts (which in general occur earlier than odd
> counts) are on the (top
> side of the) convex hull either.
>
> similarly the ?last occurrence? sequence contains points not on the
> (bottom side of the) convex
> hull of Goldbach?s Comet.
>
> [One might ask, what is the best way to put points on the convex hull
> of Goldbach?s Comet into OEIS,
> if you have a suggestion feel free to comment, those are what I
> consider the true upper and lower
> bounds, but those are pairs of integers. Yes they are integers, yes
> they can be sequenced, but they
> are pairs. Would you create two (related) sequences ? I?m not
> proposing anything on that topic,
> instead...]
>
>
> but even though the above sequences aren?t the lower and upper bounds
> I was looking for,
> and one might call them longitudinal studies of Goldbach?s Comet
> instead,
> they still seem to be of sufficient mathematical interest to be in
> OEIS, because
>
> There are some obvious conjectures
> 1. every count is the Goldbach count for some N
> equivalently every number occurs in A002372 (there is a first
> occurrence)
> 2. every count is the Goldbach count for only finitely many N,
> equivalently there is a last occurrence in A002372 for each number
>
>
> so I hope folks on this list can help with viable descriptions for
> these sequences
> such that they will be accepted into the OEIS
>
>
> what would you name these sequences
>
> (I consider them well defined, IMHO, but) how would you describe these
> sequences
>
> if you don?t think they are well defined please state why
>
>
>
> Peter A Lawrence
>
>
>
>> On Apr 18, 2021, at 6:51 PM, Neil Sloane <njasloane at gmail.com> wrote:
>>
>> Peter Lawrence,
>> Following the advice of some senior editors - but based primarily on
>> my own views - your sequence was rejected by me on April 2 2021 as
>> "interesting but not ready for the OEIS". You would have a received a
>> copy of this decision since you were the author.
>>
>> Furthermore, this was also recorded in the webpage on the OEIS Wiki
>> called Deleted Sequences.
>>
>> The reasons are documented in the Pink Box discussions of the
>> sequence, which you can see by going to the "history" tab of A342302.
>>
>> Best regards
>> Neil
>>
>> Neil J. A. Sloane, President, OEIS Foundation.
>> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
>> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway,
>> NJ.
>> Phone: 732 828 6098; home page: http://NeilSloane.com
>> <http://neilsloane.com/>
>> Email: njasloane at gmail.com <mailto:njasloane at gmail.com>
>>
>
>
>
> ------------------------------
>
> Message: 20
> Date: Wed, 21 Apr 2021 14:47:35 +0200
> From: "Richard J. Mathar" <mathar at mpia-hd.mpg.de>
> To: seqfan at list.seqfan.eu
> Subject: [seqfan] Re: Anyone recognize this matrix?
> Message-ID: <20210421124735.GA20127 at mathar.mpia-hd.mpg.de>
> Content-Type: text/plain; charset=us-ascii
>
> A formal description of this infinite array of 0's and 1's is:
> The "full" array including a leading row of all-0 starts as follows:
>
> 0 00000000000000000
> 1 01010101010101010
> 2 01000100010001000
> 3 00010001000100010
> 4 00010000000100000
> 5 01000101010001010
> 6 01010100010101000
> 7 00000001000000010
> 8 00000001000000000
> 9 01010100010101010
> 10 01000101010001000
> 11 00010000000100010
> 12 00010001000100000
> 13 01000100010001010
> 14 01010101010101000
> 15 00000000000000010
> 16 00000000000000010
> 17 01010101010101000
> 18 01000100010001010
> 19 00010001000100000
> 20 00010000000100010
> 21 01000101010001000
> 22 01010100010101010
> 23 00000001000000000
> 24 00000001000000010
> 25 01010100010101000
> 26 01000101010001010
> 27 00010000000100000
> 28 00010001000100010
> 29 01000100010001000
> 30 01010101010101010
> 31 00000000000000000
>
> Because each second column contains only zeros, we delete each second
> column
> and get the "reduced" array
>
> 0 00000000000000000
> 1 11111111111111111
> 2 10101010101010101
> 3 01010101010101010
> 4 01000100010001000
> 5 10111011101110111
> 6 11101110111011101
> 7 00010001000100010
> 8 00010000000100000
> 9 11101111111011111
> 10 10111010101110101
> 11 01000101010001010
> 12 01010100010101000
> 13 10101011101010111
> 14 11111110111111101
> 15 00000001000000010
> 16 00000001000000000
> 17 11111110111111111
> 18 10101011101010101
> 19 01010100010101010
> 20 01000101010001000
> 21 10111010101110111
> 22 11101111111011101
> 23 00010000000100010
> 24 00010001000100000
> 25 11101110111011111
> 26 10111011101110101
> 27 01000100010001010
> 28 01010101010101000
> 29 10101010101010111
> 30 11111111111111101
> 31 00000000000000010
>
> Each odd-numbered row is the binary complement of its preceding row, so
> define a "depleted reduced" array just containing rows 0,2,4,6,8,...:
>
> 0 00000000000000000
>
> 2 10101010101010101
>
> 4 01000100010001000
> 6 11101110111011101
>
> 8 00010000000100000
> 10 10111010101110101
> 12 01010100010101000
> 14 11111110111111101
>
> 16 00000001000000000
> 18 10101011101010101
> 20 01000101010001000
> 22 11101111111011101
> 24 00010001000100000
> 26 10111011101110101
> 28 01010101010101000
> 30 11111111111111101
>
> The definition of this seems to be given by reading the 1st, 2nd, 3rd
> ... column downwards, which gives periodic patterns of zeros and ones:
>
> 0,1 (col 1)
> 0,0,1,1 (col 2)
> 0,1 (col 3)
> 0,0,0,0,1,1,1,1 (col 4)
> 0,1 (col 5)
> 0,0,1,1 (col 6)
> 0,1 (col 7)
> 0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1 (col 8)
> 0,1 (col 9)
> 0,0,1,1 (col 10)
> 0,1 (col 11)
> 0,0,0,0,1,1,1,1 (col 12)
> 0,1 (col 13)
> 0,0,1,1 (col 14)
> 0,1 (col 15)
> 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1, (col
> 16)
>
> where the number of zeros (and number of ones) in the periods
> of column k is given by A006519(k).
>
>
> ------------------------------
>
> Message: 21
> Date: Wed, 21 Apr 2021 09:07:21 -0400
> From: Neil Sloane <njasloane at gmail.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Subject: [seqfan] Re: Anyone recognize this matrix?
> Message-ID:
> <CAAOnSgR1XuMEHXrhCRSXK7ujKXfTMMjzyZUKQL1_gm3SwDo__Q at mail.gmail.com>
> Content-Type: text/plain; charset="UTF-8"
>
> Richard, You are right, and indeed one can say much more. In fact this
> is
> part of a bigger investigation and there is a paper in progress that
> will
> reveal everything. I was hoping to have it finished a week ago, but
> keeping
> the OEIS running takes a great deal of time. Once the paper is in
> readable
> form I will post a link to it here.
>
> Best regards
> Neil
>
> Neil J. A. Sloane, President, OEIS Foundation.
> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway,
> NJ.
> Phone: 732 828 6098; home page: http://NeilSloane.com
> Email: njasloane at gmail.com
>
>
>
> On Wed, Apr 21, 2021 at 8:47 AM Richard J. Mathar
> <mathar at mpia-hd.mpg.de>
> wrote:
>
>> A formal description of this infinite array of 0's and 1's is:
>> The "full" array including a leading row of all-0 starts as follows:
>>
>> 0 00000000000000000
>> 1 01010101010101010
>> 2 01000100010001000
>> 3 00010001000100010
>> 4 00010000000100000
>> 5 01000101010001010
>> 6 01010100010101000
>> 7 00000001000000010
>> 8 00000001000000000
>> 9 01010100010101010
>> 10 01000101010001000
>> 11 00010000000100010
>> 12 00010001000100000
>> 13 01000100010001010
>> 14 01010101010101000
>> 15 00000000000000010
>> 16 00000000000000010
>> 17 01010101010101000
>> 18 01000100010001010
>> 19 00010001000100000
>> 20 00010000000100010
>> 21 01000101010001000
>> 22 01010100010101010
>> 23 00000001000000000
>> 24 00000001000000010
>> 25 01010100010101000
>> 26 01000101010001010
>> 27 00010000000100000
>> 28 00010001000100010
>> 29 01000100010001000
>> 30 01010101010101010
>> 31 00000000000000000
>>
>> Because each second column contains only zeros, we delete each second
>> column
>> and get the "reduced" array
>>
>> 0 00000000000000000
>> 1 11111111111111111
>> 2 10101010101010101
>> 3 01010101010101010
>> 4 01000100010001000
>> 5 10111011101110111
>> 6 11101110111011101
>> 7 00010001000100010
>> 8 00010000000100000
>> 9 11101111111011111
>> 10 10111010101110101
>> 11 01000101010001010
>> 12 01010100010101000
>> 13 10101011101010111
>> 14 11111110111111101
>> 15 00000001000000010
>> 16 00000001000000000
>> 17 11111110111111111
>> 18 10101011101010101
>> 19 01010100010101010
>> 20 01000101010001000
>> 21 10111010101110111
>> 22 11101111111011101
>> 23 00010000000100010
>> 24 00010001000100000
>> 25 11101110111011111
>> 26 10111011101110101
>> 27 01000100010001010
>> 28 01010101010101000
>> 29 10101010101010111
>> 30 11111111111111101
>> 31 00000000000000010
>>
>> Each odd-numbered row is the binary complement of its preceding row,
>> so
>> define a "depleted reduced" array just containing rows 0,2,4,6,8,...:
>>
>> 0 00000000000000000
>>
>> 2 10101010101010101
>>
>> 4 01000100010001000
>> 6 11101110111011101
>>
>> 8 00010000000100000
>> 10 10111010101110101
>> 12 01010100010101000
>> 14 11111110111111101
>>
>> 16 00000001000000000
>> 18 10101011101010101
>> 20 01000101010001000
>> 22 11101111111011101
>> 24 00010001000100000
>> 26 10111011101110101
>> 28 01010101010101000
>> 30 11111111111111101
>>
>> The definition of this seems to be given by reading the 1st, 2nd, 3rd
>> ... column downwards, which gives periodic patterns of zeros and ones:
>>
>> 0,1 (col 1)
>> 0,0,1,1 (col 2)
>> 0,1 (col 3)
>> 0,0,0,0,1,1,1,1 (col 4)
>> 0,1 (col 5)
>> 0,0,1,1 (col 6)
>> 0,1 (col 7)
>> 0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1 (col 8)
>> 0,1 (col 9)
>> 0,0,1,1 (col 10)
>> 0,1 (col 11)
>> 0,0,0,0,1,1,1,1 (col 12)
>> 0,1 (col 13)
>> 0,0,1,1 (col 14)
>> 0,1 (col 15)
>> 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1, (col
>> 16)
>>
>> where the number of zeros (and number of ones) in the periods
>> of column k is given by A006519(k).
>>
>> --
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>
>
> ------------------------------
>
> Subject: Digest Footer
>
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>
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