[seqfan] Re: Question from Harvey Dale about A233552

[Andrew Weimholt]

(419*4^n+1)/3
for n = 0 mod 6, is divisible by 5 and 7
for n = 1 mod 6, is divisible by 13
for n = 2 mod 6, is divisible by 3 and 5
for n = 3 mod 6, is divisible by 7
for n = 4 mod 6, is divisible by 5
for n = 5 mod 6, is divisible by 3

the "mod 6" arises from the fact that 4^6 - 1 = (63)(65) = 3*3*5*7*13
so
4^6 == 1 mod 5
4^6 == 1 mod 7
4^6 == 1 mod 9
4^6 == 1 mod 13

Andrew








On Sun, May 26, 2019 at 1:30 PM jean-paul allouche <
jean-paul.allouche at imj-prg.fr> wrote:

> Right!
>
> My (less than) two pence:
>
> If n is even, 2^{2^n} = 4^n is congruent to 1 modulo 5,
> so that 419. 2^{2^n} + 1 is congruent to 0 mod 5, so is its quotient by 3.
>
> Also if n is divisible by 3, 4^n is congruent to 1 modulo 7, so that
> 419. 4^n + 1 is congruent to 0 modulo 7, same for its quotient by 3.
>
> The remaining cases are when n is congruent to \pm 1 modulo 6...
> Not sure my trivial remarks are the right attack
>
> best
> jean-paul
>
>
>
> Le 26/05/2019 à 22:21, Fred Lunnon a écrit :
> > For     "(419*22n + 1)/3 can never be prime"
> > read   "(419*2^(2n) + 1)/3 can never be prime"
> >
> > WFL
> >
> >
> >
> > On 5/26/19, Neil Sloane <njasloane at gmail.com> wrote:
> >> Don't much like that idea.  Look at the link in A233551, which has a
> claim
> >> by Wesolowski that
> >> (419*22n + 1)/3 can never be prime
> >> <https://primes.utm.edu/glossary/xpage/Prime.html>. [Wesolowski
> >> <https://primes.utm.edu/curios/ByOne.php?submitter=Wesolowski>]
> >> What is the proof?
> >> We need to find a number-theorist who can straighten this out.
> >>
> >> Adding a bound on k is not an acceptable solution, imho!
> >>
> >> 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, May 26, 2019 at 3:26 PM Hugo Pfoertner <yae9911 at gmail.com>
> wrote:
> >>
> >>> http://oeis.org/A233551 shows a similar deficiency, i.e., there are
> >>> candidate terms missing in the sequence passing a test deliberately
> >>> truncated at k=10000
> >>> 2495, 3419, 3719, 5459, 5837,....
> >>> One could modify the definition of A233551 and A233552 by introducing
> an
> >>> upper limit for k, e.g. 1<=k<=n, and then add all missing terms.
> >>> A233552 would become
> >>> 25, 49, 121, 169, 289, 361, 373, 499, 529, 625, 751, 841, 919, 961,
> 1159,
> >>> 1171, 1189, 1225, 1369, 1681, 1849, 2209, 2401, 2419, 2629, 2809, 3025,
> >>> 3061, 3145, 3301, 3445, 3481, 3721, 3943, 3991, 4159, 4225, 4489, 5041,
> >>> 5209, 5329, 5461, 5539, 5581,
> >>>
> >>> A233551 would become
> >>> 89, 419, 659, 839, 1769, 2495, 2609, 2651, 2981, 3419, 3719, 4889,
> 5459,
> >>> 5561, 5771, 5837, 6341, 6509, 6971, 7271, 7829, 8447, 8609, 9521,
> >>> with 89 and 839 not passing a "for all k" condition.
> >>>
> >>>
> >>> On Sun, May 26, 2019 at 7:31 PM Neil Sloane <njasloane at gmail.com>
> wrote:
> >>>
> >>>> Harvey just asked me the following question.  Can anyone help?
> >>>>
> >>>>
> >>>> I may be missing something, but there seem to be many terms missing
> from
> >>>> the above sequence. My calculations show that, up to 1000, each of 25,
> >>> 49,
> >>>> 121, 169, 289, 361, 373, 499, 529, 613, 625, 751, 841, 919, and 961
> >>>> satisfies the definition, but only 361 and 919 appear in the data. Am
> I
> >>>> overlooking something? Also, I’m not sure how to test “all k >=1”
> >>>> because
> >>>> that would require going up to infinity — so, is there some top limit
> to
> >>>> the value of k that should be tested, e.g., k<=n? Or is there some
> other
> >>>> way to do the test that doesn’t require generating lots of terms?
> >>>>
> >>>> --
> >>>> Seqfan Mailing list - http://list.seqfan.eu/
> >>>>
> >>> --
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> >>>
> >> --
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> >>
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