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

[israel at math.ubc.ca]

There are no more primes p < 1200000 with p^2 | a(N(p)).  

Heuristically, we might guess that the probability of p^2 | a(N(p)) is 
about 1/p, and since sum_p 1/p = infinity we should expect it to occur for 
infinitely many p. But the sum grows very slowly (like log(log(n)), so it 
wouldn't be at all surprising for the next such p to be around 10^15.

Cheers,
Robert

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

>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
>>>>
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