From djr at nk.ca Tue Apr 6 10:19:20 2021
From: djr at nk.ca (Don Reble)
Date: Tue, 6 Apr 2021 02:19:20 -0600
Subject: [seqfan] A339950, A189378
Message-ID: <606C1988.90801@nk.ca>
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
From shallit at uwaterloo.ca Tue Apr 6 19:10:45 2021
From: shallit at uwaterloo.ca (Jeffrey Shallit)
Date: Tue, 6 Apr 2021 13:10:45 -0400
Subject: [seqfan] Re: A339950, A189378
In-Reply-To: <606C1988.90801@nk.ca>
References: <606C1988.90801@nk.ca>
Message-ID: <9ad00bb5-2f24-5cd5-3014-591013221ab3@uwaterloo.ca>
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?
>
From alonso.delarte at gmail.com Thu Apr 8 23:12:24 2021
From: alonso.delarte at gmail.com (Alonso Del Arte)
Date: Thu, 8 Apr 2021 17:12:24 -0400
Subject: [seqfan] How to define analogue to this sequence pertaining to
Roman numerals
Message-ID:
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
Musician at ReverbNation.com
From njasloane at gmail.com Sat Apr 10 04:06:31 2021
From: njasloane at gmail.com (Neil Sloane)
Date: Fri, 9 Apr 2021 22:06:31 -0400
Subject: [seqfan] Anyone recognize this matrix?
Message-ID:
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.]
From olivier.gerard at gmail.com Sat Apr 17 16:55:15 2021
From: olivier.gerard at gmail.com (Olivier Gerard)
Date: Sat, 17 Apr 2021 17:55:15 +0300
Subject: [seqfan] Recent perturbations on the Seqfan Mailing List
In-Reply-To:
References:
Message-ID:
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
From alonso.delarte at gmail.com Thu Apr 15 22:56:06 2021
From: alonso.delarte at gmail.com (Alonso Del Arte)
Date: Thu, 15 Apr 2021 16:56:06 -0400
Subject: [seqfan] Conjecture: a(n) = n only if n = 1 or 9
Message-ID:
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
Musician at ReverbNation.com
From tomaszordowski at gmail.com Sat Apr 17 17:45:25 2021
From: tomaszordowski at gmail.com (Tomasz Ordowski)
Date: Sat, 17 Apr 2021 17:45:25 +0200
Subject: [seqfan] Lucasian (pseudo)primes
Message-ID:
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
From dsm054 at gmail.com Sat Apr 17 20:50:57 2021
From: dsm054 at gmail.com (D. S. McNeil)
Date: Sat, 17 Apr 2021 14:50:57 -0400
Subject: [seqfan] Re: Conjecture: a(n) = n only if n = 1 or 9
In-Reply-To:
References:
Message-ID:
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
From franktaw at netscape.net Sat Apr 17 19:37:37 2021
From: franktaw at netscape.net (Frank Adams-watters)
Date: Sat, 17 Apr 2021 17:37:37 +0000 (UTC)
Subject: [seqfan] Sum-Product Problem
References: <499880394.1809449.1618681057196.ref@mail.yahoo.com>
Message-ID: <499880394.1809449.1618681057196@mail.yahoo.com>
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
From acwacw at gmail.com Thu Apr 15 03:08:03 2021
From: acwacw at gmail.com (Allan Wechsler)
Date: Wed, 14 Apr 2021 21:08:03 -0400
Subject: [seqfan] Planar distributive lattices
Message-ID:
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!
From njasloane at gmail.com Sun Apr 18 03:23:05 2021
From: njasloane at gmail.com (Neil Sloane)
Date: Sat, 17 Apr 2021 21:23:05 -0400
Subject: [seqfan] Re: Sum-Product Problem
In-Reply-To: <499880394.1809449.1618681057196@mail.yahoo.com>
References: <499880394.1809449.1618681057196.ref@mail.yahoo.com>
<499880394.1809449.1618681057196@mail.yahoo.com>
Message-ID:
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/
>
From njasloane at gmail.com Sun Apr 18 03:23:05 2021
From: njasloane at gmail.com (Neil Sloane)
Date: Sat, 17 Apr 2021 21:23:05 -0400
Subject: [seqfan] Re: Sum-Product Problem
In-Reply-To: <499880394.1809449.1618681057196@mail.yahoo.com>
References: <499880394.1809449.1618681057196.ref@mail.yahoo.com>
<499880394.1809449.1618681057196@mail.yahoo.com>
Message-ID:
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/
>
From njasloane at gmail.com Sun Apr 18 07:17:36 2021
From: njasloane at gmail.com (Neil Sloane)
Date: Sun, 18 Apr 2021 01:17:36 -0400
Subject: [seqfan] Re: Planar distributive lattices
In-Reply-To:
References:
Message-ID:
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 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/
>
From njasloane at gmail.com Sun Apr 18 07:38:57 2021
From: njasloane at gmail.com (Neil Sloane)
Date: Sun, 18 Apr 2021 01:38:57 -0400
Subject: [seqfan] Re: Planar distributive lattices
In-Reply-To:
References:
Message-ID:
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 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 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/
>>
>
From yae9911 at gmail.com Sun Apr 18 11:10:42 2021
From: yae9911 at gmail.com (Hugo Pfoertner)
Date: Sun, 18 Apr 2021 11:10:42 +0200
Subject: [seqfan] Re: Sum-Product Problem
In-Reply-To:
References: <499880394.1809449.1618681057196.ref@mail.yahoo.com>
<499880394.1809449.1618681057196@mail.yahoo.com>
Message-ID:
A263996 is even more closely related. I also
added the link there.
On Sun, Apr 18, 2021 at 3:23 AM Neil Sloane 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/
>
From alonso.delarte at gmail.com Sun Apr 18 20:05:21 2021
From: alonso.delarte at gmail.com (Alonso Del Arte)
Date: Sun, 18 Apr 2021 14:05:21 -0400
Subject: [seqfan] Re: Conjecture: a(n) = n only if n = 1 or 9
In-Reply-To:
References:
Message-ID:
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 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
Musician at ReverbNation.com
From acwacw at gmail.com Sun Apr 18 23:36:45 2021
From: acwacw at gmail.com (Allan Wechsler)
Date: Sun, 18 Apr 2021 17:36:45 -0400
Subject: [seqfan] Re: Planar distributive lattices
In-Reply-To:
References:
Message-ID:
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 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 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 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/
>
From njasloane at gmail.com Mon Apr 19 04:10:06 2021
From: njasloane at gmail.com (Neil Sloane)
Date: Sun, 18 Apr 2021 22:10:06 -0400
Subject: [seqfan] Re: Planar distributive lattices
In-Reply-To:
References:
Message-ID:
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 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 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 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
> 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/
>
From peterl95124 at comcast.net Mon Apr 19 06:54:25 2021
From: peterl95124 at comcast.net (peter lawrence)
Date: Sun, 18 Apr 2021 21:54:25 -0700
Subject: [seqfan] help naming/describing sequences for bounds of Goldbach's
Comet, was: what happened to my A342302, its been replaced ???
In-Reply-To:
References:
Message-ID: <5B52ED1E-B02E-42B9-A71C-7E5AA36BCEE5@comcast.net>
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 , 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 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
> Email: njasloane at gmail.com
>
From mathar at mpia-hd.mpg.de Wed Apr 21 14:47:35 2021
From: mathar at mpia-hd.mpg.de (Richard J. Mathar)
Date: Wed, 21 Apr 2021 14:47:35 +0200
Subject: [seqfan] Re: Anyone recognize this matrix?
Message-ID: <20210421124735.GA20127@mathar.mpia-hd.mpg.de>
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).
From njasloane at gmail.com Wed Apr 21 15:07:21 2021
From: njasloane at gmail.com (Neil Sloane)
Date: Wed, 21 Apr 2021 09:07:21 -0400
Subject: [seqfan] Re: Anyone recognize this matrix?
In-Reply-To: <20210421124735.GA20127@mathar.mpia-hd.mpg.de>
References: <20210421124735.GA20127@mathar.mpia-hd.mpg.de>
Message-ID:
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
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/
>
From david.j.seal at gwynmop.com Wed Apr 21 12:24:44 2021
From: david.j.seal at gwynmop.com (David Seal)
Date: Wed, 21 Apr 2021 11:24:44 +0100 (BST)
Subject: [seqfan] Re: Planar distributive lattices
In-Reply-To:
References:
Message-ID: <2088376732.13963.1619000684306@email.ionos.co.uk>
With regard to question (a), I think the Wikipedia pages https://en.wikipedia.org/wiki/Distributive_lattice and https://en.wikipedia.org/wiki/Hasse_diagram contain the required information. My only uncertainty is whether 'planar' is being used as shorthand for 'that has an upward planar Hasse diagram', but the poster clearly shows Hasse diagrams that are upward planar, and the diagrams do not include one of the subsets of a 3-element set {x,y,z} ordered by subset inclusion. That lattice has a Hasse diagram which is the skeleton of a cube (see https://en.wikipedia.org/wiki/Partially_ordered_set#/media/File:Hasse_diagram_of_powerset_of_3.svg), and the skeleton of a cube is of course a planar graph - but drawing it in a planar fashion necessarily causes it to violate the 'upward' property of Hasse diagrams and so it isn't upward planar. So the fact that the poster's diagrams do not include the skeleton of a cube tends to support my belief that in 'planar distributive lattice', 'planar' is referring to upward planarity of the Hasse diagram and not just planarity of a diagram.
With regard to question (c), I've checked Dr. Jipsen's poster for numbering the diagrams correctly and the diagrams being listed in non-decreasing order of their vertex count (so that all listed diagrams with 1 vertex come before all listed diagrams with 2 vertices, which come before all listed diagrams with 3 vertices, etc). That means that the sequence can be determined by a(n) = (number of first diagram with n+1 vertices) - (number of first diagram with n vertices). Doing that does indeed confirm your counts.
With regard to question (d), the decision seems to have already been made to include it in the OEIS - and for what little it's worth, I completely agree with that decision.
With regard to question (b), I'm afraid I cannot currently confirm whether Dr. Jipson has enumerated them correctly, but I didn't notice any obvious problems such as duplicates or clear omissions.
David
> On 15/04/2021 02:08 Allan Wechsler 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!
>
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