OFFSET
0,3
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
FORMULA
E.g.f.: 2*exp(x)/(1 + exp(x) + exp(2*x) - exp(3*x)).
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 *)
PROG
(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])
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Prabha Sivaramannair, Feb 09 2024
STATUS
approved