A305291 Numbers k such that 4*k + 3 is a perfect cube, sorted by absolute values.

-1, 6, -32, 85, -183, 332, -550, 843, -1229, 1714, -2316, 3041, -3907, 4920, -6098, 7447, -8985, 10718, -12664, 14829, -17231, 19876, -22782, 25955, -29413, 33162, -37220, 41593, -46299, 51344, -56746, 62511, -68657, 75190, -82128, 89477, -97255, 105468, -114134, 123259, -132861

Offset

1,2

Links

Colin Barker, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (-3,-2,2,3,1).

Formula

G.f.: x*(-1 + 3*x - 16*x^2 + 3*x^3 - x^4)/((1 - x)*(1 + x)^4).

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

a(n) = (-3 + A016755(n-1)*(-1)^n)/4.

a(n) = -A305290(n) - 1.

a(n) + a(-n) = 1 - 2^(1+(-1)^n).

(n - 2)*(4*n^2 - 16*n + 19)*a(n) + (12*n^2 - 36*n + 31)*a(n-1) - (n - 1)*(4*n^2 - 8*n + 7)*a(n-2) = 0.

From Colin Barker, May 30 2018: (Start)

a(n) = (4*n^3 - 6*n^2 + 3*n - 2)/2 for n even.

a(n) = -(4*n^3 - 6*n^2 + 3*n + 1)/2 for n odd.

(End)

Maple

seq(coeff(series(x*(-1+3*x-16*x^2+3*x^3-x^4)/((1-x)*(1+x)^4), x, 50), x, n), n=1..45); # Muniru A Asiru, May 31 2018

Mathematica

LinearRecurrence[{-3, -2, 2, 3, 1}, {-1, 6, -32, 85, -183}, 45] (* Jean-François Alcover, Jun 04 2018 *)

Prog

(PARI) Vec(-x*(1 - 3*x + 16*x^2 - 3*x^3 + x^4) / ((1 - x)*(1 + x)^4) + O(x^40)) \\ Colin Barker, Jun 04 2018

Crossrefs

Cf. A016755, A305290.

Sequence in context: A045159 A121002 A161844 * A177082 A296196 A211918

Adjacent sequences: A305288 A305289 A305290 * A305292 A305293 A305294

Keyword

sign,easy

Author

Bruno Berselli, May 29 2018

Status

approved