[seqfan] Re: Question from Harvey Dale about A233552

[jean-paul allouche]

Excellent!
jpa

Le 26/05/2019 à 23:33, Andrew Weimholt a écrit :
> (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
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> 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?
>>>>>>
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>>>>>>
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