A320627 - OEIS

%I #11 Oct 24 2018 13:33:14

%S 1,1,1,13,217,12938197,913083596083,3836387399699939518675459471,

%T 18744974860247264575032720770000376335095039,

%U 25741458812593689971179132474269180614331431944325835714919500509967358371226305360396760987

%N a(n) = A006134(A038754(n) - 1)/3^n.

%C a(n) is always an integer. k = A038754(n) - 1 is the smallest index that A006134(k) is divisible by 3^n.

%C The next term has 140 digits.

%C For primes p we have A006134(p-1) == Legendre(p, 3) (mod p^2). For composite n that is a power of 3, n^2 is also divisible by A006134(n-1). Are there any other such n?

%C Conjecture: for n > 1, a(n) == 1 (mod 27) for even n, a(n) == 13 (mod 27) for odd n.

%e a(1) = (binomial(0, 0) + binomial(2, 1))/3 = 3/3 = 1.

%e a(2) = (binomial(0, 0) + binomial(2, 1) + binomial(4, 2))/9 = 9/9 = 1.

%e a(3) = (binomial(0, 0) + binomial(2, 1) + binomial(4, 2) + binomial(6, 3) + binomial(8, 4) + binomial(10, 5))/27 = 351/27 = 13.

%t Array[Sum[Binomial[2 k, k], {k, 0, #}] &[((1 + Boole[OddQ@ #]) 3^((# - Boole[OddQ@ #])/2)) - 1]/3^# &, 9] (* _Michael De Vlieger_, Oct 22 2018 *)

%o (PARI) A006134(n) = sum(k=0,n,binomial(2*k,k))

%o a(n) = if(n%2, A006134(2*3^((n-1)/2)-1)/3^n, A006134(3^(n/2)-1)/3^n)

%Y Cf. A000244, A006134, A038754, A320625.

%K nonn

%O 0,4

%A _Jianing Song_, Oct 18 2018