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

[israel at math.ubc.ca]

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
>>
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