A275638 Expansion of (1-q)^k/Product_{j=1..k} (1-q^j) for k=4.

1, -3, 4, -3, 2, -3, 5, -6, 6, -6, 6, -6, 7, -9, 10, -9, 8, -9, 11, -12, 12, -12, 12, -12, 13, -15, 16, -15, 14, -15, 17, -18, 18, -18, 18, -18, 19, -21, 22, -21, 20, -21, 23, -24, 24, -24, 24, -24, 25, -27, 28, -27, 26, -27, 29, -30, 30, -30, 30, -30, 31, -33, 34, -33, 32, -33, 35, -36, 36

Offset: 0

Author

N. J. A. Sloane, Aug 09 2016

Keywords

• sign
• easy

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• A. M. Odlyzko, Differences of the partition function, Acta Arithmetica 49.3 (1988): 237-254.
• Dennis Stanton and Doron Zeilberger, The Odlyzko conjecture and O’Hara’s unimodality proof, Proceedings of the American Mathematical Society 107.1 (1989): 39-42.
• Index entries for linear recurrences with constant coefficients, signature (-3,-5,-6,-5,-3,-1)

Crossrefs

Cf. A275639, A275640, A275641, A275642, A275643, A275644.

Programs

Maple

f1:=k->(1-q)^k/mul(1-q^i,i=1..k);
f2:=k->series(f1(k),q,75);
f3:=k->seriestolist(f2(k));
f3(4);

PARI

Vec(1/((1+x)^2*(1+x^2)*(1+x+x^2)) + O(x^100)) \\ Colin Barker, Aug 11 2016

Formula

Equivalent g.f.: 1 / ((1+x)^2*(1+x^2)*(1+x+x^2)). - Colin Barker, Aug 10 2016

From Ilya Gutkovskiy, Aug 10 2016: (Start)

a(n) = -3*a(n-1) - 5*a(n-2) - 6*a(n-3) - 5*a(n-4) - 3*a(n-5) - a(n-6).

a(n) = (sqrt(3)*(-1)^n*n + 3*sqrt(3)*(-1)^n - 4*sin(2*Pi n/3) - sqrt(3)*cos(Pi*n/2))/(2*sqrt(3)). (End)