OFFSET
1,1
COMMENTS
Numbers k such that k^3 + (k+1)^3 + ... + (k+M-1)^3 = c^2 has nontrivial solutions over the integers where M is an odd positive integer.
To every odd positive integer M corresponds a sum of M consecutive cubes starting at a(n) having at least one nontrivial solution. For n >= 1, M(n) = (2n+1) (A005408), a(n) = M^3 - (3M-1)/2 = (2n+1)^3 - (3n+1) and c(n) = M*(M^2-1)*(2M^2-1)/2 = 2n*(n+1)*(2n+1)*(8n*(n+1)+1) (A253680).
The trivial solutions with M < 1 and k < 2 are not considered here.
Stroeker stated that all odd values of M yield a solution to k^3 + (k+1)^3 + ... + (k+M-1)^3 = c^2. This was further demonstrated by Pletser.
LINKS
Vladimir Pletser, Table of n, a(n) for n = 1..50000
Vladimir Pletser, File Triplets (M,a,c) for M=(2n+1)
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) = (2n+1)^3 - (3n+1).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Colin Barker, Jan 09 2015
G.f.: -x*(x^2-26*x-23) / (x-1)^4. - Colin Barker, Jan 09 2015
EXAMPLE
For n=1, M(n)=3, a(n)=23, c(n)=204.
See "File Triplets (M,a,c) for M=(2n+1)" link.
MAPLE
for n from 1 to 50 do a:=(2*n+1)^3-(3*n+1): print (a); end do:
MATHEMATICA
a253679[n_] := (2 # + 1)^3 - (3 # + 1) & /@ Range@ n; a253679[52] (* Michael De Vlieger, Jan 10 2015 *)
PROG
(PARI) Vec(-x*(x^2-26*x-23)/(x-1)^4 + O(x^100)) \\ Colin Barker, Jan 09 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vladimir Pletser, Jan 08 2015
STATUS
approved