A320625 a(n) = A006134((3^n-1)/2)/3^n.

1, 1, 11, 520783, 1777232132705889910073, 1989655738014873996462170980393276816167557169374094238588991602837393

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.

Links

Table of n, a(n) for n=0..5.

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

Cf. A000244, A006134, A320627.

Sequence in context: A076171 A297057 A090102 * A219013 A243130 A253632

Adjacent sequences: A320622 A320623 A320624 * A320626 A320627 A320628

Keyword

nonn

Author

Jianing Song, Oct 18 2018

Status

approved