A253709 Integer squares c^2 that are equal to the sums of M (A253707) 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).

104329, 56205009, 3319833924, 68869504900, 771665618025, 5755695204609, 32148582480784, 144648440352144, 550265331330225, 1830621686635225, 5457952678249764, 14853496612506084, 37420748658691489, 88243404864147225, 196505988636801600, 416206765369428544, 843426135281228409, 1643334148974958209, 3091319880732100900, 5634162244739340900

Offset

1,1

Comments

Numbers a(n)=c^2 such that b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2 has nontrivial solutions over the integers for b being an odd squared integer (A016754) and M (A253707).

To every odd squared integer b (A016754) corresponds a sum of M (A253707) consecutive cubed integers starting at b^3 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) (A253707), c(n)= (b-1)(4b^2-1)/8 = (n(n+1)/2)(4(2n+1)^4-1) (A253708) and a(n)=c(n)^2 (this sequence).

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, 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 (13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1).

Formula

a(n) = ((n(n+1)/2)(4(2n+1)^4-1))^2.

G.f.: -x*(104329*x^10 +54848732*x^9 +2597306469*x^8 +30065816496*x^7 +119309063058*x^6 +186443360232*x^5 +119309063058*x^4 +30065816496*x^3 +2597306469*x^2 +54848732*x +104329) / (x -1)^13. - Colin Barker, Jan 10 2015

Example

For n=1, b(1)=9, M(1)=17, c(1)=323, a(1)= 104329 (see File File Triplets (M,b,c) for a=(2n+1)^2 link).

Maple

restart: for n from 1 to 50000 do a:=((n*(n+1)/2)(4*(2*n+1)^4-1))^2: print (a); end do:

Mathematica

f[n_] := ((n (n + 1)/2) (4 (2 n + 1)^4 - 1))^2; Array[f, 20] (* Michael De Vlieger, Jan 10 2015 *)

Prog

(PARI) Vec(-x*(104329*x^10 +54848732*x^9 +2597306469*x^8 +30065816496*x^7 +119309063058*x^6 +186443360232*x^5 +119309063058*x^4 +30065816496*x^3 +2597306469*x^2 +54848732*x +104329) / (x -1)^13 + O(x^100)) \\ Colin Barker, Jan 10 2015
(Magma) [((n*(n+1)/2)*(4*(2*n+1)^4-1))^2: n in [1..20]]; // Vincenzo Librandi, Jan 15 2015

Crossrefs

Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253679, A253680, A253707, A253708.

Sequence in context: A236007 A129241 A290535 * A235005 A231143 A249528

Adjacent sequences: A253706 A253707 A253708 * A253710 A253711 A253712

Keyword

nonn,easy

Author

Vladimir Pletser, Jan 09 2015

Status

approved