[seqfan] Re: Question from Harvey Dale about A233552
jean-paul allouche
jean-paul.allouche at imj-prg.fr
Sun May 26 22:30:30 CEST 2019
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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