A130168 a(n) = (b(n) + b(n+1))/3, where b(n) = A000366(n).

1, 3, 15, 111, 1131, 15123, 256335, 5364471, 135751731, 4084163643, 144039790455, 5884504366431, 275643776229531, 14673941326078563, 880908054392169375, 59226468571935857991, 4432461082611507366531, 367227420727722013775883, 33514867695588319595233095

Offset: 2

Author

Don Knuth, Aug 02 2007

Keywords

• nonn

Comments

As remarked by Gessel, A000366 has a combinatorial interpretation via a certain 2n X n array; this sequence is for a similar array of size (2n-1) X (n-1).

In effect, Dellac gives a combinatorial reason why the elements of A000366 are alternately -1 and +1 modulo 3. Dellac also shows that all the terms of this sequence are odd.

Links

• Hippolyte Dellac, Note sur l'élimination, méthode de parallélogramme, Annales de la Faculté des Sciences de Marseille, XI (1901), 141-164. [Warning 76 Mb; go to p. 81 in the pdf file]

Crossrefs

Cf. A000366, A130169.

Programs

Mathematica

b[n_] := (-2^(-1))^(n-2)*Sum[Binomial[n, k]*(1-2^(n+k+1))* BernoulliB[n+k+1], {k, 0, n}];
a[n_] := (b[n] + b[n+1])/3;
a /@ Range[2, 20] (* Jean-François Alcover, Apr 08 2021 *)

Python

from math import comb
from sympy import bernoulli
def A130168(n): return (abs((2-(2<>n-1)//3 # Chai Wah Wu, Apr 14 2023

Formula

G.f.: 2*(1+x)/(3*x^3)*Q(0) - 2/(3*x) - 1/x^2 - 2/(3*x^3), where Q(k) = 1 - x*(k+1)^2/( x*(k+1)^2 - 2/(1 - x*(k+1)^2/( x*(k+1)^2 - 2/Q(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Oct 22 2013

a(n) = |(2-2^(n+2))*Bernoulli(n+1) - (n+1)*(1-2^(2n+2))*Bernoulli(2n+2) - (1-2^(2n+3))*Bernoulli(2n+3) + Sum_{k=0..n-1} (2*binomial(n,k+1)-binomial(n+1,k))*(1-2^(n+k+2))*Bernoulli(n+k+2)|/(3*2^(n-1)). - Chai Wah Wu, Apr 14 2023