A180920 Numbers k such that the sum of the cubes of the k consecutive integers starting from k is a square.

1, 33, 2017, 124993, 7747521, 480221281, 29765971873, 1845010034817, 114360856186753, 7088528073543841, 439374379703531361, 27234123013545400513, 1688076252460111300417, 104633493529513355225313, 6485588522577367912668961, 402001854906267297230250241

Offset: 1

Author

Vladimir Pletser, Sep 24 2010

Keywords

• nonn
• easy

Comments

Numbers k such that A240137(k) is a square.

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

Links

• Colin Barker, Table of n, a(n) for n = 1..558
• Vladimir Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015.
• R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
• Index entries for linear recurrences with constant coefficients, signature (63,-63,1).

Crossrefs

Cf. A180921, A240137.

Programs

Mathematica

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 *)

PARI

default(realprecision, 1000)
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

Formula

a(n) = 31*a(n-1) - 14 + 8*sqrt((3*a(n-1) - 1)*(5*a(n-1) - 3)).

From Colin Barker, Feb 18 2015: (Start)

a(n) = 63*a(n-1) - 63*a(n-2) + a(n-3).

G.f.: -x*(x^2-30*x+1) / ((x-1)*(x^2-62*x+1)). (End)

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

Extensions

a(8) onwards from Robert G. Wilson v, Sep 27 2010

Name clarified by Jon E. Schoenfield, Mar 11 2022