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A262018 The first of five consecutive positive integers the sum of the squares of which is equal to the sum of the squares of eleven consecutive positive integers.

%I A262018 #7 May 16 2019 11:31:45
%S A262018 28,5308,945148,168231388,29944242268,5329906892668,948693482652988,
%T A262018 168862110005339548,30056506887467786908,5349889363859260730428,
%U A262018 952250250260060942229628,169495194656926988456143708,30169192398682743884251350748,5369946751770871484408284289788
%N A262018 The first of five consecutive positive integers the sum of the squares of which is equal to the sum of the squares of eleven consecutive positive integers.
%C A262018 For the first of the corresponding eleven consecutive positive integers, see A262019.
%H A262018 Colin Barker, <a href="/A262018/b262018.txt">Table of n, a(n) for n = 1..444</a>
%H A262018 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (179,-179,1).
%F A262018 a(n) = 179*a(n-1)-179*a(n-2)+a(n-3) for n>3.
%F A262018 G.f.: -4*x*(7*x^2+74*x+7) / ((x-1)*(x^2-178*x+1)).
%e A262018 28 is in the sequence because 28^2 + ... + 32^2 = 4510 = 15^2 + ... + 25^2.
%t A262018 LinearRecurrence[{179,-179,1},{28,5308,945148},30] (* _Harvey P. Dale_, May 16 2019 *)
%o A262018 (PARI) Vec(-4*x*(7*x^2+74*x+7)/((x-1)*(x^2-178*x+1)) + O(x^20))
%Y A262018 Cf. A157096, A262017, A262019.
%K A262018 nonn,easy
%O A262018 1,1
%A A262018 _Colin Barker_, Sep 08 2015