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
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).
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
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