Number factorization and a highly suspicous claim

[Max]

Oh, the same actually follows from the formula
A001333(n) = ((1-sqrt(2))^n + (1+sqrt(2))^n)/2
which does not require any knowledge of Chebyshev's polynomials.

Max

Max wrote:
> Creighton Dement wrote:
> 
>> %S A111108
>> 0,5,3,25,25,131,175,705,1137,3875,7095,21649,43225,122435,259423,
>> %T A111108
>> 698625,1541985,4011971,9107175,23143825,53559817,133933475,314086735,
>> %U A111108
>> 776787009,1838300625,4512108515,10745077143,26237143825,62749602745
>> %N A111108 a(n) = A001333(n) - (-2)^(n-1), n > 0.
>> %C A111108 Conjecture: for prime p, p divides a(p). Note that (a(n)) and
>> A001333 have different offsets.
>> %D A111108 C. Dement, Floretion Integer Sequences (work in progress)
>> %F A111108 G.f. -x*(3*x+5)/((2*x+1)*(x^2+2*x-1))
>> %Y A111108 Cf. A001333, A000204.
>> %K A111108 easy,nonn,new
>> %O A111108 1,2
>> %A A111108 Creighton Dement (crowdog(AT)crowdog.de), Oct 14 2005
> 
> There is a very simple proof of the conjecture taking into account the 
> formula
> A001333(2n+1) = S(2n,2*sqrt(2)) = U(2n,sqrt(2)),
> where U(n,x), Chebyshev's polynomials of the second kind.
> 
> Let p=2n+1 be an odd prime.
> 
> According to formula (14) at 
> http://mathworld.wolfram.com/ChebyshevPolynomialoftheSecondKind.html
> U(2n,sqrt(2)) = \sum C(2n+1,2m+1) 2^(n-m)
> Taking it modulo prime p=2n+1 we have
> U(2n,sqrt(2)) = C(p,p) 2^0 = 1 (mod p)
> 
> Therefore,
> A111108(p) = A001333(p) - (-2)^(p-1) = 1 - 1 = 0 (mod p)
> 
> Max
> 
>