A370092 a(0) = 1, a(n) = (-1)^n + (1/2) * Sum_{j=1..n} (1-(-1)^j-(-2)^j) * binomial(n,j) * a(n-j) for n > 0.

1, 1, 3, 16, 105, 856, 8433, 96916, 1272225, 18789136, 308335713, 5565837916, 109603592145, 2338198823416, 53718370204593, 1322292130204516, 34718481333932865, 968552056638097696, 28609403248435931073, 892022330159009036716, 29276492753074019702385

Offset: 0

Author

Prabha Sivaramannair, Feb 09 2024

Keywords

• nonn

Comments

Inverse binomial transform of A370456.

Conjecture: Let k > 2 be a positive integer. The sequence obtained by reducing a(n) modulo k is eventually periodic with the period dividing phi(k) = A000010(k). For example, modulo 10 we obtain the sequence [1, 1, 3, 6, 5, 6, 3, 6, 5, 6, 3, 6, 5, 6, 3, 6, 5, 6, ...] with an apparent period of 4 beginning at a(2). See A000670 for a more general conjecture. - Peter Bala, Feb 16 2024

Crossrefs

Cf. A370163, A370456.

Programs

Mathematica

a[0]=1;Table[(-1)^n+Sum[ (1-(-1)^j-  (-2) ^j) *Binomial[n,j]*a[n-j]/2,{j,1,n} ],{n,0,20}] (* James C. McMahon, Feb 10 2024 *)

PARI

seq(n)={my(p=exp(x + O(x*x^n))); Vec(serlaplace(2*p/(1 + p + p^2 - p^3)))} \\ Andrew Howroyd, Feb 10 2024

SageMath

def a(m):
    if m==0:
        return 1
    else:
        return (-1)^m+1/2*sum([(1-(-2)^j-(-1)^j)*binomial(m,j)*a(m-j) for j in [1,..,m]])
list(a(m) for m in [0,..,20])

Formula

E.g.f.: 2*exp(x)/(1 + exp(x) + exp(2*x) - exp(3*x)).