A101918 G.f. satisfies: A(x) = 1/(1 + x*A(x^8)) and also the continued fraction: 1 + x*A(x^9) = [1; 1/x, 1/x^8, 1/x^64, 1/x^512, ..., 1/x^(8^(n-1)), ...].

1, -1, 1, -1, 1, -1, 1, -1, 1, 0, -1, 2, -3, 4, -5, 6, -7, 7, -6, 4, -1, -3, 8, -14, 21, -28, 34, -38, 39, -36, 28, -14, -7, 35, -69, 107, -146, 182, -210, 224, -217, 182, -113, 6, 140, -322, 532, -756, 973, -1155, 1268, -1274, 1134, -812, 280, 476, -1449, 2604, -3872, 5146, -6280, 7092, -7372, 6896, -5447, 2843, 1029

Offset

0,12

Links

Paul D. Hanna, Table of n, a(n) for n = 0..1000

Formula

The conjecture that this has g.f. (1+x^8) / (1+x+x^8) by Ralf Stephan, May 17 2007 is wrong. The first difference occurs at a(73) = -42106. The g.f. gives a(73) = -42105. - Johannes W. Meijer, Aug 08 2011

a(0) = 1; a(n) = -Sum_{k=0..floor((n-1)/8)} a(k) * a(n-8*k-1). - Ilya Gutkovskiy, Mar 01 2022

Maple

nmax:=66: kmax:=nmax: for k from 0 to kmax do A:= proc(x): add(A101918(n)*x^n, n=0..k) end: f(x):=series(1/(1 + x*A(x^8)), x, k+1); for n from 0 to k do x(n):=coeff(f(x), x, n) od: A101918(k):=x(k): od: seq(A101918(n), n=0..nmax); # Johannes W. Meijer, Aug 08 2011

Mathematica

nmax = 66; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x] - (1/(1 + x A[x^8])) + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
sol /. Rule -> Set;
a /@ Range[0, nmax] (* Jean-François Alcover, Nov 03 2019 *)

Prog

(PARI) {a(n)=local(A); A=1-x; for(i=1, n\8+1, A=1/(1+x*subst(A, x, x^8)+x*O(x^n))); polcoeff(A, n, x)}
for(n=0, 120, print1(a(n), ", "))
(PARI) {a(n)=local(M=contfracpnqn(concat(1, vector(ceil(log(n+1)/log(8))+1, n, 1/x^(8^(n-1)))))); polcoeff(M[1, 1]/M[2, 1]+x*O(x^(9*n+1)), 9*n+1)}
for(n=0, 120, print1(a(n), ", "))

Crossrefs

Cf. A101912-A101917.

Sequence in context: A264856 A141258 A117656 * A291169 A323633 A363275

Adjacent sequences: A101915 A101916 A101917 * A101919 A101920 A101921

Keyword

sign

Author

Paul D. Hanna, Dec 20 2004

Status

approved