A253679 Numbers that begin a run of an odd number of consecutive integers whose cubes sum to a square.

23, 118, 333, 716, 1315, 2178, 3353, 4888, 6831, 9230, 12133, 15588, 19643, 24346, 29745, 35888, 42823, 50598, 59261, 68860, 79443, 91058, 103753, 117576, 132575, 148798, 166293, 185108, 205291, 226890, 249953, 274528, 300663, 328406, 357805, 388908, 421763, 456418, 492921, 531320, 571663, 613998, 658373, 704836, 753435, 804218, 857233, 912528, 970151, 1030150, 1092573, 1157468

Offset: 1

Author

Vladimir Pletser, Jan 08 2015

Keywords

• nonn
• easy

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.

Examples

			For n=1, M(n)=3, a(n)=23, c(n)=204.
See "File Triplets (M,a,c) for M=(2n+1)" link.
		

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).

Crossrefs

Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253680, A253681.

Programs

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 *)

PARI

Vec(-x*(x^2-26*x-23)/(x-1)^4 + O(x^100)) \\ Colin Barker, Jan 09 2015

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