OFFSET
0,3
COMMENTS
a(n) is always an integer.
a(6) has 215 digits and a(7) has 654 digits.
For primes p we have A006134((p-1)/2) == Legendre(p, 3) (mod p). For composite n equal to a power of 3, n is also divisible by A006134((n-1)/2). Odd composite n not a power of 3 such that n divides A006134((n-1)/2) are n = 99, 1539, ... What's the next? Are there similar values of n not divisible by 3?
Conjecture: for n > 0, a(n) == 1 (mod 3) for odd n, a(n) == 2 (mod 3) for even n.
EXAMPLE
a(1) = (binomial(0, 0) + binomial(2, 1))/3 = 3/3 = 1.
a(2) = (binomial(0, 0) + binomial(2, 1) + binomial(4, 2) + binomial(6, 3) + binomial(8, 4))/9 = 99/9 = 11.
MAPLE
a:=n->add(binomial(2*k, k), k=0..(3^n-1)/2)/3^n: seq(a(n), n=0..5); # Muniru A Asiru, Oct 23 2018
MATHEMATICA
Array[Sum[Binomial[2 k, k], {k, 0, #}] &[(3^# - 1)/2]/3^# &, 5] (* Michael De Vlieger, Oct 22 2018 *)
PROG
(PARI) A006134(n) = sum(k=0, n, binomial(2*k, k))
a(n) = A006134((3^n-1)/2)/3^n
(GAP) List([0..5], n->Sum([0..(3^n-1)/2], k->Binomial(2*k, k))/3^n); # Muniru A Asiru, Oct 23 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, Oct 18 2018
STATUS
approved