A261972 -id:A261972 - cp's OEIS frontend

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

Colin Barker, Sep 07 2015

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)

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

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

Cf. A157088, A261972, A261974.

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

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

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

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

A261974 The first of eleven consecutive positive integers the sum of the squares of which is equal to the sum of the squares of three consecutive positive integers.

Original entry on oeis.org

67, 3307, 152275, 7001563, 321919843, 14801311435, 680538406387, 31289965382587, 1438657869192835, 66146972017488043, 3041322054935257363, 139834667555004350875, 6429353385475264883107, 295610421064307180272267, 13591650015572655027641395

Offset: 1

Colin Barker, Sep 07 2015

For the first of the corresponding three consecutive positive integers, see A261973.

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

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

Cf. A157088, A261972, A261973.

LinearRecurrence[{47,-47,1},{67,3307,152275},20] (* Harvey P. Dale, Jul 02 2016 *)

Vec(x*(5*x^2-158*x-67)/((x-1)*(x^2-46*x+1)) + O(x^40))

a(n) = 47*a(n-1)-47*a(n-2)+a(n-3) for n>3.
G.f.: x*(5*x^2-158*x-67) / ((x-1)*(x^2-46*x+1)).
a(n) = -5+3*sqrt(3/11)*(23+4*sqrt(33))^(-n)*(-1+(23+4*sqrt(33))^(2*n)). - Colin Barker, Mar 03 2016

cp's OEIS Frontend

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.

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