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A180920
Numbers k such that the sum of the cubes of the k consecutive integers starting from k is a square.
3
1, 33, 2017, 124993, 7747521, 480221281, 29765971873, 1845010034817, 114360856186753, 7088528073543841, 439374379703531361, 27234123013545400513, 1688076252460111300417, 104633493529513355225313, 6485588522577367912668961, 402001854906267297230250241
OFFSET
1,2
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
R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
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
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 *)
PROG
(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
CROSSREFS
Sequence in context: A330478 A284221 A264170 * A120288 A284112 A099370
KEYWORD
nonn,easy
AUTHOR
Vladimir Pletser, Sep 24 2010
EXTENSIONS
a(8) onwards from Robert G. Wilson v, Sep 27 2010
Name clarified by Jon E. Schoenfield, Mar 11 2022
STATUS
approved