A251867 Numbers n such that n^2 + (n+1)^2 is equal to the sum of the hexagonal numbers H(m) and H(m+1) for some m.

0, 14, 492, 16730, 568344, 19306982, 655869060, 22280241074, 756872327472, 25711378892990, 873430010034204, 29670908962269962, 1007937474707144520, 34240203231080643734, 1163158972382034742452, 39513164857758100599650, 1342284446191393385645664

Offset: 1

Author

Colin Barker, Dec 10 2014

Keywords

• nonn
• easy

Comments

Also nonnegative integers y in the solutions to 4*x^2-2*y^2+2*x-2*y = 0, the corresponding values of x being A220185.

Examples

			14 is in the sequence because 14^2+15^2 = 196+225 = 421 = 190+231 = H(10)+H(11).
		

Links

• Colin Barker, Table of n, a(n) for n = 1..654
• Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Crossrefs

Cf. A000290, A000384, A220185, A105635.

Programs

Magma

I:=[0,14,492]; [n le 3 select I[n] else 35*Self(n-1)-35*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Sep 06 2015

Mathematica

LinearRecurrence[{35, -35, 1}, {0, 14, 492}, 20] (* Vincenzo Librandi, Sep 06 2015 *)

PARI

concat(0, Vec(-2*x^2*(x+7)/((x-1)*(x^2-34*x+1)) + O(x^100)))

Formula

a(n) = 35*a(n-1)-35*a(n-2)+a(n-3).

G.f.: -2*x^2*(x+7) / ((x-1)*(x^2-34*x+1)).

a(n) = A220185(n) + A001542(n-1)^2. - Alexander Samokrutov, Sep 05 2015

a(n) = (-4+(10+7*sqrt(2))*(17+12*sqrt(2))^(-n)+(10-7*sqrt(2))*(17+12*sqrt(2))^n)/8. - Colin Barker, Mar 02 2016

a(n) = A105635(4*n-4). - Greg Dresden, Aug 30 2021