A101912 G.f. satisfies: A(x) = 1/(1 + x*A(x^2)) and also the continued fraction: 1 + x*A(x^3) = [1; 1/x, 1/x^2, 1/x^4, 1/x^8, ..., 1/x^(2^(n-1)), ...].

1, -1, 1, 0, -1, 1, 0, -2, 3, -1, -3, 6, -4, -4, 12, -10, -5, 23, -25, -2, 43, -57, 12, 74, -124, 56, 120, -258, 172, 170, -516, 454, 187, -989, 1095, 40, -1811, 2487, -604, -3128, 5375, -2567, -4991, 11140, -7704, -6976, 22164, -20062, -7220, 42288, -48020, -36, 76928, -108334, 29476, 131898, -233020, 117166

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. - Johannes W. Meijer, Aug 08 2011

Links

Table of n, a(n) for n=0..57.

Formula

G.f.: 1/(1 + x/(1 + x^2/(1 + x^4/(1 + x^8/(1 + ...))))) (continued fraction). - Joerg Arndt, Oct 19 2012

G.f. A(x) = 1/B(x) where B(x) is the g.f. of A218031. - Joerg Arndt, Oct 19 2012

a(0) = 1; a(n) = -Sum_{k=0..floor((n-1)/2)} a(k) * a(n-2*k-1). - Ilya Gutkovskiy, Mar 01 2022

Maple

nmax:=57: kmax:=nmax: for k from 0 to kmax do A:= proc(x): add(A101912(n)*x^n, n=0..k) end: f(x):=series(1/(1 + x*A(x^2)), x, k+1); for n from 0 to k do x(n):=coeff(f(x), x, n) od: A101912(k):=x(k): od: seq(A101912(n), n=0..nmax); # Johannes W. Meijer, Aug 08 2011

Mathematica

m = 58; A[_] = 0; Do[A[x_] = 1/(1 + x A[x^2]) + 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\2+1, A=1/(1+x*subst(A, x, x^2)+x*O(x^n))); polcoeff(A, n, x)}
(PARI) {a(n)=local(M=contfracpnqn(concat(1, vector(#binary(n)+1, n, 1/x^(2^(n-1)))))); polcoeff(M[1, 1]/M[2, 1]+x*O(x^(3*n+1)), 3*n+1)}

Crossrefs

Cf. A101913, A101914, A101915, A101916, A101917, A101918.

Sequence in context: A118981 A117938 A256193 * A208522 A209569 A368158

Adjacent sequences: A101909 A101910 A101911 * A101913 A101914 A101915

Keyword

sign

Author

Paul D. Hanna, Dec 20 2004

Status

approved