A253707 - OEIS

A253707

Numbers M(n) which are the number of terms in the sums of consecutive cubed integers equaling a squared integer, b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2, for a first term b(n) being an odd squared integer (A016754).

17, 98, 291, 644, 1205, 2022, 3143, 4616, 6489, 8810, 11627, 14988, 18941, 23534, 28815, 34832, 41633, 49266, 57779, 67220, 77637, 89078, 101591, 115224, 130025, 146042, 163323, 181916, 201869, 223230, 246047, 270368, 296241, 323714, 352835, 383652, 416213

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OFFSET

1,1

COMMENTS

Numbers M(n) such that b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2 has nontrivial solutions over the integers for b(n) being an odd squared integer (A016754).

To every odd squared integer b corresponds a sum of a consecutive cubed integers starting at b having at least one nontrivial solution. For n>=1, b(n)= (2n+1)^2 (A016754), M(n) = (sqrt(b)-1) (2b-1)/2 = n(8n(n+1)+1) (this sequence), and c(n)= (b-1)(4b^2-1)/8 = (n (n+1)/2)(4(2n+1)^4-1) (A253708).

The trivial solutions with M < 1 and b < 2 are not considered here.

LINKS

Vladimir Pletser, Table of n, a(n) for n = 1..50000

Vladimir Pletser, File Triplets (M,b,c) for a=(2n+1)^2

Vladimir Pletser, Number of terms, first term and square root of sums of consecutive cubed integers equal to integer squares, Research Gate, 2015.

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 (4,-6,4,-1).

FORMULA

a(n) = n(8n(n+1)+1).

a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Colin Barker, Jan 10 2015

G.f.: x*(x^2+30*x+17) / (x-1)^4. - Colin Barker, Jan 10 2015