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%I #23 Sep 08 2022 08:46:13
%S 137,6341,291593,13406981,616429577,28342353605,1303131836297,
%T 59915722116101,2754820085504393,126661808211086021,
%U 5823688357624452617,267763002642513734405,12311274433198007330057,566050860924465823448261,26026028328092229871289993
%N 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.
%C For the first of the corresponding eleven consecutive positive integers, see A261974.
%C From _Zak Seidov_, Sep 07 2015: (Start)
%C 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.
%C 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)
%H Colin Barker, <a href="/A261973/b261973.txt">Table of n, a(n) for n = 1..601</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (47,-47,1).
%F a(n) = 47*a(n-1)-47*a(n-2)+a(n-3) for n>3.
%F G.f.: -x*(5*x^2-98*x+137) / ((x-1)*(x^2-46*x+1)).
%F a(n) = -1+3*(23+4*sqrt(33))^(-n)+3*(23+4*sqrt(33))^n. - _Colin Barker_, Mar 03 2016
%e 137 is in the sequence because 137^2 + 138^2 + 139^2 = 57134 = 67^2 + ... + 77^2.
%t LinearRecurrence[{47, -47, 1}, {137, 6341, 291593}, 20] (* _Vincenzo Librandi_, Sep 08 2015 *)
%o (PARI) Vec(-x*(5*x^2-98*x+137) / ((x-1)*(x^2-46*x+1)) + O(x^40))
%o (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
%Y Cf. A157088, A261972, A261974.
%K nonn,easy
%O 1,1
%A _Colin Barker_, Sep 07 2015
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