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Numbers n such that the hexagonal number H(n) is equal to the sum of four consecutive squares.
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%I #6 Jun 13 2015 00:55:20

%S 18,42,602,1418,20442,48162,694418,1636082,23589762,55578618,

%T 801357482,1888036922,27222564618,64137676722,924765839522,

%U 2178792971618,31414815979122,74014823358282,1067178977450618,2514325201209962,36252670417341882,85413042017780418

%N Numbers n such that the hexagonal number H(n) is equal to the sum of four consecutive squares.

%C Also positive integers y in the solutions to 8*x^2-4*y^2+24*x+2*y+28 = 0, the corresponding values of x being A201633.

%H Colin Barker, <a href="/A252747/b252747.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,34,-34,-1,1).

%F a(n) = a(n-1)+34*a(n-2)-34*a(n-3)-a(n-4)+a(n-5).

%F G.f.: -2*x*(x^4-26*x^2+12*x+9) / ((x-1)*(x^2-6*x+1)*(x^2+6*x+1)).

%e 18 is in the sequence because H(18) = 630 = 121+144+169+196 = 11^2+12^2+13^2+14^2.

%o (PARI) Vec(-2*x*(x^4-26*x^2+12*x+9)/((x-1)*(x^2-6*x+1)*(x^2+6*x+1)) + O(x^100))

%Y Cf. A000290, A000384, A201633,

%K nonn,easy

%O 1,1

%A _Colin Barker_, Dec 21 2014