A101914 G.f. satisfies: A(x) = 1/(1 + x*A(x^4)) and also the continued fraction: 1 + x*A(x^5) = [1; 1/x, 1/x^4, 1/x^16, 1/x^64, ..., 1/x^(4^(n-1)), ...].

1, -1, 1, -1, 1, 0, -1, 2, -3, 3, -2, 0, 3, -6, 8, -8, 5, 1, -9, 17, -22, 20, -10, -8, 31, -51, 60, -50, 16, 38, -100, 150, -163, 119, -11, -147, 315, -432, 433, -268, -70, 522, -964, 1222, -1118, 542, 484, -1756, 2887, -3385, 2793, -879, -2176, 5678, -8472, 9186, -6672, 542, 8372, -17816, 24384, -24350, 14952

Offset

0,8

Comments

Sequence appears to have a rational g.f. - Ralf Stephan, May 17 2007

The conjecture is wrong. The g.f. is dependent on the number of terms. With twenty terms the g.f. is (1 + x^4)/(1 + x + x^4). - Johannes W. Meijer, Aug 08 2011

Links

Table of n, a(n) for n=0..62.

Formula

a(0) = 1; a(n) = -Sum_{k=0..floor((n-1)/4)} a(k) * a(n-4*k-1). - Ilya Gutkovskiy, Mar 01 2022

Maple

nmax:=62: kmax:=nmax: for k from 0 to kmax do A:= proc(x): add(A101914(n)*x^n, n=0..k) end: f(x):=series(1/(1 + x*A(x^4)), x, k+1); for n from 0 to k do x(n):=coeff(f(x), x, n) od: A101914(k):=x(k): od: seq(A101914(n), n=0..nmax); # Johannes W. Meijer, Aug 08 2011

Mathematica

m = 63; A[_] = 0; Do[A[x_] = 1/(1 + x A[x^4]) + O[x]^m // Normal, {m}]; CoefficientList[A[x], x] (* Jean-François Alcover, Nov 03 2019 *)

Prog

(PARI) a(n)=local(A); A=1-x; for(i=1, n\4+1, A=1/(1+x*subst(A, x, x^4)+x*O(x^n))); polcoeff(A, n, x)
(PARI) a(n)=local(M=contfracpnqn(concat(1, vector(ceil(log(n+1)/log(4))+1, n, 1/x^(4^(n-1)))))); polcoeff(M[1, 1]/M[2, 1]+x*O(x^(5*n+1)), 5*n+1)

Crossrefs

Cf. A101912, A101913, A101915, A101916, A101917, A101918.

Sequence in context: A323186 A022906 A285408 * A099530 A167544 A334484

Adjacent sequences: A101911 A101912 A101913 * A101915 A101916 A101917

Keyword

sign

Author

Paul D. Hanna, Dec 20 2004

Status

approved