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%I #41 Apr 14 2019 09:00:50
%S 0,1,47,2161,99359,4568353,210044879,9657496081,444034774847,
%T 20415942146881,938689303981679,43159292041010353,1984388744582494559,
%U 91238722958753739361,4194996867358089516047,192878617175513363998801,8868221393206256654428799,407745305470312292739725953,18747415830241159209372965039
%N Integers m such that m^3 is the sum of squares of m consecutive integers.
%H Ion Cucurezeanu and Gertrude Ehrlich, <a href="https://www.jstor.org/stable/2322431">Cubes as sums of consecutive squares: Problem E3064</a>, Amer. Math. Monthly, Vol. 94, No. 2 (Feb., 1987), pp. 190-192.
%H Philippe Fondanaiche, <a href="http://people.missouristate.edu/lesreid/Sol02_1112_Fondanaiche.pdf">Challenge problem 2011-2012 #02</a>, solution to Missouri State University's Challenge Problem.
%H Problems in Elementary Number Theory (PEN), <a href="http://www.artofproblemsolving.com/Forum/viewtopic.php?f=470&t=150917">Problem P 15</a>, Art of Problem Solving website, 2007.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (46, -1).
%F For n>3, a(n) = 46*a(n-1) - a(n-2).
%F G.f.: x^2*(1+x)/(1-46*x+x^2). - _Bruno Berselli_, Apr 19 2011
%F a(n) = ((-517+90*sqrt(33))*(23+4*sqrt(33))^n-(517+90*sqrt(33))*(23-4*sqrt(33))^n)/22 for n>1, a(1)=0. - _Bruno Berselli_, May 31 2011
%t LinearRecurrence[{46, -1}, {0, 1, 47}, 19] (* a(1) prepended by _Georg Fischer_, Apr 03 2019 *)
%o (Maxima) makelist(if n=1 then 0 else expand(((90*sqrt(33)-517)*(23+4*sqrt(33))^n-(90*sqrt(33)+517)*(23-4*sqrt(33))^n)/22),n,1,19); /* _Bruno Berselli_, May 31 2011 */
%o (PARI) x='x+O('x^19); Vec(x^2*(1+x)/(1-46*x+x^2)) \\ _Georg Fischer_, Apr 03 2019
%K nonn,easy
%O 1,3
%A _Max Alekseyev_, Apr 18 2011
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