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A262019 The first of eleven consecutive positive integers the sum of the squares of which is equal to the sum of the squares of five consecutive positive integers.

%I A262019 #5 Sep 09 2015 03:28:45
%S A262019 15,3575,637215,113421575,20188404015,3593422493975,639609015524415,
%T A262019 113846811340852775,20264092809656270415,3606894673307475281975,
%U A262019 642006987755920943922015,114273636925880620542837575,20340065365818994535681167215,3620417361478855146730704927575
%N A262019 The first of eleven consecutive positive integers the sum of the squares of which is equal to the sum of the squares of five consecutive positive integers.
%C A262019 For the first of the corresponding five consecutive positive integers, see A262018.
%H A262019 Colin Barker, <a href="/A262019/b262019.txt">Table of n, a(n) for n = 1..444</a>
%H A262019 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (179,-179,1).
%F A262019 a(n) = 179*a(n-1)-179*a(n-2)+a(n-3) for n>3.
%F A262019 G.f.: 5*x*(5*x^2-178*x-3) / ((x-1)*(x^2-178*x+1)).
%e A262019 15 is in the sequence because 15^2 + ... + 25^2 = 4510 = 28^2 + ... + 32^2.
%o A262019 (PARI) Vec(5*x*(5*x^2-178*x-3)/((x-1)*(x^2-178*x+1)) + O(x^20))
%Y A262019 Cf. A157096, A262017, A262018.
%K A262019 nonn,easy
%O A262019 1,1
%A A262019 _Colin Barker_, Sep 08 2015