[seqfan] Re: R: On the divisibility of a certain linear recurrence

[israel at math.ubc.ca]

I might also add: N(2) is undefined, N(41)=41, otherwise N(p) is a divisor 
of p-1 if 41 is a quadratic residue mod p, and a divisor of p+1 if 41 is 
not a quadratic residue mod p.

Cheers,
Robert

On Jan 9 2019, israel at math.ubc.ca wrote:

> For prime p, let N(p) be the least n > 0 such that a(n) == 0 mod p. Then 
> a(n) == 0 mod p iff n is divisible by N(p). So you're looking for cases 
> where N(p) is prime and p^2 | a(N(p)). We have N(5) = 4, with 5^2 | a(4) 
> = 75 (but 4 is not prime), and N(19) = 5, with a(5) = 19^2 (and 5 is 
> prime). The next time p^2 | a(N(p)) is p = 1193, with N(1193) = 596 (not 
> prime). No more such p below 20000. This is somewhat reminiscent of 
> Fibonacci-Wieferich primes (or Wall-Sun-Sun primes).
>
>Cheers,
>Robert
>
>On Jan 9 2019, Hugo Pfoertner wrote:
>
>>Hi Peter,
>>
>> I've checked through p=157 and didn't find a counterexample. Of course 
>> this is only an empirical observation. The factorization of a(163) is 
>> hard.
>>
>> Are there any other coefficients of a linear recurrence 
>> a(n)=c1*a(n-1)+c2*a(n-2), other than c1= 3 and c2=8, that produce a 
>> similar behavior for n=prime?
>>
>> If not, the usual objection against making sequences using arbitrary 
>> parameters might not apply and we could create at least 3 sequences, 
>> starting from a(2)=3 Least prime factor of a(p(n)): 3, 17, 19, 7937, 
>> 3875057, 85655881, 67, 419, 5107, 59, 28585163, 1627, 41, 563041, 283, 
>> 107, 6435362107, 1925667763, 4153, 2131, 
>> 1850491543540265361576345006181354087969129016281, 514289, 331, 37201, 
>> 193, 10041419, 619, 6430415092373729, 3923, 5744017, 5843, 1571, 
>> 36537899, 377801, 1787, 2113, 313 Largest prime factor of a(p(n)): 3, 
>> 17, 19, 7937, 3875057, 85655881, 624669523, 2207981563, 88514071931, 
>> 548466097, 3775503987139, 716442143452957321483, 1780775573699, 
>> 5811596076161, 38609629351582573090820353, 1817374561603, 
>> 143077761236233553, 1768827268814468956771, 
>> 225877944042691675469399613458636969, 13010415851368044971, 
>> 1850491543540265361576345006181354087969129016281, 3540092609483175233, 
>> 38828807550428376855712721787851, 3715881986067409049, ... number of 
>> prime factors of a(p(n)): 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 2, 2, 5, 3, 3, 
>> 6, 3, 3, 3, 5, 1, 4, 3, 6, 3, 3, 2, 3, 3, 5, 5, 5, 2, 2, 4, 7, 4,
>>
>>Hugo Pfoertner
>>
>>BTW: Tomorrow Don Knuth will celebrate his 3^4-th birthday ...
>>
>>
>>
>>
>>On Tue, Jan 8, 2019 at 10:44 AM Mason John <john.mason at lispa.it> wrote:
>>
>>> a(3)=3*a(2) + 8*a(1) = 3*3 + 8*1 = 17, right? Why 9?
>>> john
>>>
>>> -----Messaggio originale-----
>>> Da: SeqFan <seqfan-bounces at list.seqfan.eu> Per conto di Peter Luschny
>>> Inviato: martedì 8 gennaio 2019 10:30
>>> A: seqfan at list.seqfan.eu
>>> Oggetto: [seqfan] On the divisibility of a certain linear recurrence
>>>
>>> Consider the linear recurrence
>>>
>>>     a(n) = 3*a(n-1) + 8*a(n-2) starting 1, 3, 9, ...
>>>
>>> Is it true that p prime and p not 2 or 5 implies that a(p) is 
>>> squarefree?
>>>
>>> Peter
>>>
>>> --
>>> Seqfan Mailing list - http://list.seqfan.eu/
>>>
>>> --
>>> Seqfan Mailing list - http://list.seqfan.eu/
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>>
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