A320627 a(n) = A006134(A038754(n) - 1)/3^n.

1, 1, 1, 13, 217, 12938197, 913083596083, 3836387399699939518675459471, 18744974860247264575032720770000376335095039, 25741458812593689971179132474269180614331431944325835714919500509967358371226305360396760987

Offset

0,4

Comments

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

The next term has 140 digits.

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?

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

Links

Table of n, a(n) for n=0..9.

Example

a(1) = (binomial(0, 0) + binomial(2, 1))/3 = 3/3 = 1.
a(2) = (binomial(0, 0) + binomial(2, 1) + binomial(4, 2))/9 = 9/9 = 1.
a(3) = (binomial(0, 0) + binomial(2, 1) + binomial(4, 2) + binomial(6, 3) + binomial(8, 4) + binomial(10, 5))/27 = 351/27 = 13.

Mathematica

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

Prog

(PARI) A006134(n) = sum(k=0, n, binomial(2*k, k))
a(n) = if(n%2, A006134(2*3^((n-1)/2)-1)/3^n, A006134(3^(n/2)-1)/3^n)

Crossrefs

Cf. A000244, A006134, A038754, A320625.

Sequence in context: A096141 A218475 A294982 * A059525 A086147 A329073

Adjacent sequences: A320624 A320625 A320626 * A320628 A320629 A320630

Keyword

nonn

Author

Jianing Song, Oct 18 2018

Status

approved