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A101913
G.f. satisfies: A(x) = 1/(1 + x*A(x^3)) and also the continued fraction: 1+x*A(x^4) = [1;1/x,1/x^3,1/x^9,1/x^27,...,1/x^(3^(n-1)),...].
6
1, -1, 1, -1, 2, -3, 4, -6, 9, -13, 19, -28, 41, -61, 90, -132, 195, -288, 424, -625, 922, -1359, 2004, -2955, 4356, -6423, 9471, -13963, 20587, -30355, 44755, -65987, 97293, -143449, 211503, -311844, 459785, -677912, 999524, -1473709, 2172854, -3203685, 4723551, -6964461, 10268490, -15139986
OFFSET
0,5
FORMULA
From Joerg Arndt, Oct 15 2011: (Start)
For the sequence abs(a(n)) we have
g.f. B(x) 1/(1-x/(1-x^3/(1-x^9/(1-x^27(1- ... ))))) and
B(x) satisfies B(x) = 1 + x*B(x)*B(x^3) (cf. A000621)
(End)
G.f.: T(0), where T(k) = 1 - (-x)^(3^k)/((-x)^(3^k) - 1/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 18 2013
a(0) = 1; a(n) = -Sum_{k=0..floor((n-1)/3)} a(k) * a(n-3*k-1). - Ilya Gutkovskiy, Mar 01 2022
PROG
(PARI) {a(n)=local(A); A=1-x; for(i=1, n\3+1, A=1/(1+x*subst(A, x, x^3)+x*O(x^n))); polcoeff(A, n, x)}
(PARI) {a(n)=local(M=contfracpnqn(concat(1, vector(ceil(log(n+1)/log(3))+1, n, 1/x^(3^(n-1)))))); polcoeff(M[1, 1]/M[2, 1]+x*O(x^(4*n+1)), 4*n+1)}
CROSSREFS
KEYWORD
sign
AUTHOR
Paul D. Hanna, Dec 20 2004
STATUS
approved