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%I #40 Jan 19 2024 11:22:37
%S 1,33,2017,124993,7747521,480221281,29765971873,1845010034817,
%T 114360856186753,7088528073543841,439374379703531361,
%U 27234123013545400513,1688076252460111300417,104633493529513355225313,6485588522577367912668961,402001854906267297230250241
%N Numbers k such that the sum of the cubes of the k consecutive integers starting from k is a square.
%C Numbers k such that A240137(k) is a square.
%C A240137(k) is a square iff (3n-1)*(5n-3) is a square. 15*n^2-14*n+3 = k^2 can be solved as a Pell equation resulting in formula below confirming linear recurrence from Colin Barker. - _Ray Chandler_, Jan 12 2024
%H Colin Barker, <a href="/A180920/b180920.txt">Table of n, a(n) for n = 1..558</a>
%H Vladimir Pletser, <a href="http://arxiv.org/abs/1501.06098">General solutions of sums of consecutive cubed integers equal to squared integers</a>, arXiv:1501.06098 [math.NT], 2015.
%H R. J. Stroeker, <a href="http://www.numdam.org/item?id=CM_1995__97_1-2_295_0">On the sum of consecutive cubes being a perfect square</a>, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (63,-63,1).
%F a(n) = 31*a(n-1) - 14 + 8*sqrt((3*a(n-1) - 1)*(5*a(n-1) - 3)).
%F From _Colin Barker_, Feb 18 2015: (Start)
%F a(n) = 63*a(n-1) - 63*a(n-2) + a(n-3).
%F G.f.: -x*(x^2-30*x+1) / ((x-1)*(x^2-62*x+1)). (End)
%F a(n) = (14+2*(4*(31-8*Sqrt(15))^n+Sqrt(15)*(31-8*Sqrt(15))^n+4*(31+8*Sqrt(15))^n-Sqrt(15)*(31+8*Sqrt(15))^n))/30. - _Ray Chandler_, Jan 12 2024
%t a[1] = 1; a[n_] := a[n] = 31 a[n - 1] - 14 + 8 Sqrt[(3 a[n - 1] - 1)*(5 a[n - 1] - 3)]; Array[a, 14] (* _Robert G. Wilson v_, Sep 27 2010 *)
%o (PARI)
%o default(realprecision, 1000)
%o vector(20, n, if(n==1, t=1, t=round(31*t-14+8*((3*t-1)*(5*t-3))^(1/2)))) \\ _Colin Barker_, Feb 19 2015
%Y Cf. A180921, A240137.
%K nonn,easy
%O 1,2
%A _Vladimir Pletser_, Sep 24 2010
%E a(8) onwards from _Robert G. Wilson v_, Sep 27 2010
%E Name clarified by _Jon E. Schoenfield_, Mar 11 2022
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