A189173 Integers m such that m^3 is the sum of squares of m consecutive integers.
0, 1, 47, 2161, 99359, 4568353, 210044879, 9657496081, 444034774847, 20415942146881, 938689303981679, 43159292041010353, 1984388744582494559, 91238722958753739361, 4194996867358089516047, 192878617175513363998801, 8868221393206256654428799, 407745305470312292739725953, 18747415830241159209372965039
Offset: 1
Links
- Ion Cucurezeanu and Gertrude Ehrlich, Cubes as sums of consecutive squares: Problem E3064, Amer. Math. Monthly, Vol. 94, No. 2 (Feb., 1987), pp. 190-192.
- Philippe Fondanaiche, Challenge problem 2011-2012 #02, solution to Missouri State University's Challenge Problem.
- Problems in Elementary Number Theory (PEN), Problem P 15, Art of Problem Solving website, 2007.
- Index entries for linear recurrences with constant coefficients, signature (46,-1).
Programs
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Mathematica
LinearRecurrence[{46, -1}, {0, 1, 47}, 19] (* a(1) prepended by Georg Fischer, Apr 03 2019 *) -
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 */
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PARI
my(x='x+O('x^19)); Vec(x^2*(1+x)/(1-46*x+x^2)) \\ Georg Fischer, Apr 03 2019
Formula
For n>3, a(n) = 46*a(n-1) - a(n-2).
G.f.: x^2*(1+x)/(1-46*x+x^2). - Bruno Berselli, Apr 19 2011
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