A236343 Expansion of (1 - x + 2*x^2 - x^3) / ((1 - x)^2 * (1 - x^3)) in powers of x.

1, 1, 3, 5, 6, 9, 12, 14, 18, 22, 25, 30, 35, 39, 45, 51, 56, 63, 70, 76, 84, 92, 99, 108, 117, 125, 135, 145, 154, 165, 176, 186, 198, 210, 221, 234, 247, 259, 273, 287, 300, 315, 330, 344, 360, 376, 391, 408, 425, 441, 459, 477, 494, 513, 532, 550, 570, 590

Offset: 0

Author

Michael Somos, Jan 22 2014

Keywords

• nonn
• easy

Comments

The sequence is a quasi-polynomial sequence.

Given a sequence of Laurent polynomials defined by b(n) = (b(n-2)^2 - b(n-1)*b(n-3) * 2/x) / b(n-4), b(-2) = x, b(-4) = -b(-3) = -b(-1) = 1. Then the denominator of b(n) is x^a(n).

Examples

			G.f. = 1 + x + 3*x^2 + 5*x^3 + 6*x^4 + 9*x^5 + 12*x^6 + 14*x^7 + 18*x^8 + ...
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..2500
• Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Crossrefs

Cf. A236337. Trisections are A000326, A095794, A045943.

Programs

Magma

m:=60; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1-x+2*x^2-x^3)/((1-x)^2*(1-x^3)))); // G. C. Greubel, Aug 07 2018

Maple

seq(coeff(series((1-x+2*x^2-x^3)/((1-x)^2*(1-x^3)),x,n+1), x, n), n = 0 .. 60); # Muniru A Asiru, Feb 12 2019

Mathematica

CoefficientList[Series[(1-x+2*x^2-x^3)/((1-x)^2*(1-x^3)), {x, 0, 60}], x] (* G. C. Greubel, Aug 07 2018 *)

PARI

{a(n) = (n * (n+5) + [6, 0, 4][n%3 + 1]) / 6};

PARI

{a(n) = if( n<0, polcoeff( x^2 * (-1 + 2*x - x^2 + x^3) / ((1 - x)^2 * (1 - x^3)) + x * O(x^-n), -n), polcoeff( (1 - x + 2*x^2 - x^3) / ((1 - x)^2 * (1 - x^3)) + x * O(x^n), n))};

Sage

((1-x+2*x^2-x^3)/((1-x)^2*(1-x^3))).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Feb 12 2019

Formula

0 = a(n)*(a(n+2) + a(n+3)) + a(n+1)*(-2*a(n+2) - a(n+3) + a(n+4)) + a(n+2)*(a(n+2) - 2*a(n+3) + a(n+4)) for all n in Z.

G.f.: (1 - x + 2*x^2 - x^3) / ((1 - x)^2 * (1 - x^3)).

Second difference is period 3 sequence [2, 0, -1, ...].

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

a(-6-n) = A236337(n).

From Peter Bala, Feb 11 2019: (Start)

a(3*n) = (1/2)*(n + 1)*(3*n + 2);

a(3*n+1) = (1/2)*(n + 1)*(3*n + 4) - 1;

a(3*n+2) = (1/2)*(n + 1)*(3*n + 6). (End)