A261973 The first of three consecutive positive integers the sum of the squares of which is equal to the sum of the squares of eleven consecutive positive integers.

137, 6341, 291593, 13406981, 616429577, 28342353605, 1303131836297, 59915722116101, 2754820085504393, 126661808211086021, 5823688357624452617, 267763002642513734405, 12311274433198007330057, 566050860924465823448261, 26026028328092229871289993

Offset: 1

Author

Colin Barker, Sep 07 2015

Keywords

• nonn
• easy

Comments

For the first of the corresponding eleven consecutive positive integers, see A261974.

From Zak Seidov, Sep 07 2015: (Start)

Positive values x of solutions (x, y) to the Diophantine equation 380 + 110x + 11x^2 - 6y - 3y^2 = 0, with values of y in A261974.

Note that there are also solutions with negative x: (x,y) = (-77,137), (-3317, 6341), (-152285, 291593), (-7001573, 13406981), ... with values of y in A261974. (End)

Examples

			137 is in the sequence because 137^2 + 138^2 + 139^2 = 57134 = 67^2 + ... + 77^2.
		

Links

• Colin Barker, Table of n, a(n) for n = 1..601
• Index entries for linear recurrences with constant coefficients, signature (47,-47,1).

Crossrefs

Cf. A157088, A261972, A261974.

Programs

Magma

I:=[137,6341,291593]; [n le 3 select I[n] else 47*Self(n-1)-47*Self(n-2)+Self(n-3): n in [1..15]]; // Vincenzo Librandi, Sep 08 2015

Mathematica

LinearRecurrence[{47, -47, 1}, {137, 6341, 291593}, 20] (* Vincenzo Librandi, Sep 08 2015 *)

PARI

Vec(-x*(5*x^2-98*x+137) / ((x-1)*(x^2-46*x+1)) + O(x^40))

Formula

a(n) = 47*a(n-1)-47*a(n-2)+a(n-3) for n>3.

G.f.: -x*(5*x^2-98*x+137) / ((x-1)*(x^2-46*x+1)).

a(n) = -1+3*(23+4*sqrt(33))^(-n)+3*(23+4*sqrt(33))^n. - Colin Barker, Mar 03 2016