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A189173
Integers m such that m^3 is the sum of squares of m consecutive integers.
0
0, 1, 47, 2161, 99359, 4568353, 210044879, 9657496081, 444034774847, 20415942146881, 938689303981679, 43159292041010353, 1984388744582494559, 91238722958753739361, 4194996867358089516047, 192878617175513363998801, 8868221393206256654428799, 407745305470312292739725953, 18747415830241159209372965039
OFFSET
1,3
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.
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
MATHEMATICA
LinearRecurrence[{46, -1}, {0, 1, 47}, 19] (* a(1) prepended by Georg Fischer, Apr 03 2019 *)
PROG
(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 */
(PARI) x='x+O('x^19); Vec(x^2*(1+x)/(1-46*x+x^2)) \\ Georg Fischer, Apr 03 2019
CROSSREFS
Sequence in context: A280707 A324433 A289984 * A162896 A163265 A163803
KEYWORD
nonn,easy
AUTHOR
Max Alekseyev, Apr 18 2011
STATUS
approved