A144927 Numbers n such that there exists an integer k with (n+7)^3-n^3=k^2.

7, 1162, 128191, 14100226, 1550897047, 170584575322, 18762752388751, 2063732178187666, 226991776848254887, 24967031721129850282, 2746146497547435276511, 302051147698496750566306, 33222880100337095127017527, 3654214759889381967221362042

Offset: 1

Author

Richard Choulet, Sep 25 2008

Keywords

• easy
• nonn

Examples

			a(1)=7 because 14^3-7^3=49^2.
		

Links

• Colin Barker, Table of n, a(n) for n = 1..450
• Index entries for linear recurrences with constant coefficients, signature (111,-111,1).

Crossrefs

Cf. A144928, A144930, A144929.

Programs

Mathematica

Last /@ Table[n /. {ToRules[Reduce[n > 0 && k >= 0 && (n + 7)^3 - n^3 == k^2, n, Integers] /. C[1] -> c]} // Simplify, {c, 1, 14}] (* or *)
Rest@ CoefficientList[Series[7 x (-1 - 55 x + 2 x^2)/((x - 1) (x^2 - 110 x + 1)), {x, 0, 14}], x] (* Michael De Vlieger, Jul 14 2016 *)
LinearRecurrence[{111,-111,1},{7,1162,128191},20] (* Harvey P. Dale, Jul 05 2024 *)

PARI

Vec(7*x*(-1-55*x+2*x^2)/((x-1)*(x^2-110*x+1)) + O(x^20)) \\ Colin Barker, Jul 14 2016

Formula

a(n+2) = 110*a(n+1) - a(n) + 378.

G.f.: 7*x*(-1-55*x+2*x^2) / ( (x-1)*(x^2-110*x+1) ). - R. J. Mathar, Nov 27 2011

a(n) = 7*A144929(n). - R. J. Mathar, Nov 27 2011