A252630 Numbers n such that the sum of the hexagonal numbers X(n), X(n+1), X(n+2) and X(n+3) is equal to the heptagonal number H(m) for some m.

50, 16503, 5314316, 1711193649, 550999041062, 177419980028715, 57128682570205568, 18395258367626164581, 5923216065693054789914, 1907257177894796016188127, 614130888066058624157787380, 197748238700092982182791348633, 63674318730541874204234656472846

Offset

1,1

Comments

Also positive integers x in the solutions to 16*x^2-5*y^2+40*x+3*y+44 = 0, the corresponding values of y being A252631.

Links

Colin Barker, Table of n, a(n) for n = 1..398

Index entries for linear recurrences with constant coefficients, signature (323,-323,1).

Formula

a(n) = 323*a(n-1)-323*a(n-2)+a(n-3).

G.f.: x*(3*x^2-353*x-50) / ((x-1)*(x^2-322*x+1)).

a(n) =-5/4+1/80*(161+72*sqrt(5))^(-n)*(-70-37*sqrt(5)+(-70+37*sqrt(5))*(161+72*sqrt(5))^(2*n)). - Colin Barker, Mar 03 2016

Example

50 is in the sequence because X(50)+X(51)+X(52)+X(53) = 4950+5151+5356+5565 = 21022 = H(92).

Prog

(PARI) Vec(x*(3*x^2-353*x-50)/((x-1)*(x^2-322*x+1)) + O(x^100))

Crossrefs

Cf. A000384, A000566, A252631.

Sequence in context: A364511 A203097 A028465 * A034205 A173170 A146518

Adjacent sequences: A252627 A252628 A252629 * A252631 A252632 A252633

Keyword

nonn,easy

Author

Colin Barker, Dec 19 2014

Status

approved