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A189173 Integers m such that m^3 is the sum of squares of m consecutive integers.

%I A189173 #47 Apr 30 2025 09:14:36
%S A189173 0,1,47,2161,99359,4568353,210044879,9657496081,444034774847,
%T A189173 20415942146881,938689303981679,43159292041010353,1984388744582494559,
%U A189173 91238722958753739361,4194996867358089516047,192878617175513363998801,8868221393206256654428799,407745305470312292739725953,18747415830241159209372965039
%N A189173 Integers m such that m^3 is the sum of squares of m consecutive integers.
%H A189173 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 A189173 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 A189173 Problems in Elementary Number Theory (PEN), <a href="https://artofproblemsolving.com/community/c146h150917">Problem P 15</a>, Art of Problem Solving website, 2007.
%H A189173 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (46,-1).
%F A189173 For n>3, a(n) = 46*a(n-1) - a(n-2).
%F A189173 G.f.: x^2*(1+x)/(1-46*x+x^2). - _Bruno Berselli_, Apr 19 2011
%F A189173 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 A189173 LinearRecurrence[{46, -1}, {0, 1, 47}, 19] (* a(1) prepended by _Georg Fischer_, Apr 03 2019 *)
%o A189173 (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 A189173 (PARI) my(x='x+O('x^19)); Vec(x^2*(1+x)/(1-46*x+x^2)) \\ _Georg Fischer_, Apr 03 2019
%K A189173 nonn,easy
%O A189173 1,3
%A A189173 _Max Alekseyev_, Apr 18 2011