A305290 Numbers k such that 4*k + 1 is a perfect cube, sorted by absolute values.

0, -7, 31, -86, 182, -333, 549, -844, 1228, -1715, 2315, -3042, 3906, -4921, 6097, -7448, 8984, -10719, 12663, -14830, 17230, -19877, 22781, -25956, 29412, -33163, 37219, -41594, 46298, -51345, 56745, -62512, 68656, -75191, 82127, -89478, 97254, -105469, 114133, -123260, 132860

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^2*(-7 + 10*x - 7*x^2)/((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) = (-1 - A016755(n-1)*(-1)^n)/4.

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

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

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

(End)

Maple

seq(coeff(series(x^2*(-7+10*x-7*x^2)/((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}, {0, -7, 31, -86, 182}, 45] (* Jean-François Alcover, Jun 04 2018 *)

Prog

(PARI) concat(0, Vec(-x^2*(7 - 10*x + 7*x^2) / ((1 - x)*(1 + x)^4) + O(x^40))) \\ Colin Barker, Jun 04 2018

Crossrefs

Cf. A016755.

Cf. A000290: k such that 4*k is a square.

Cf. A002378: k such that 4*k+1 is a square.

Cf. A033431: k such that 4*k is a nonnegative cube.

Cf. A305291: k such that 4*k+3 is a cube.

Cf. A141046: k such that 4*k is a fourth power.

Cf. 4*A219086: k such that 4*k+1 is a fourth power.

Sequence in context: A201477 A164621 A202254 * A262012 A118934 A118935

Adjacent sequences: A305287 A305288 A305289 * A305291 A305292 A305293

Keyword

sign,easy

Author

Bruno Berselli, May 29 2018

Status

approved