A253725 Integer squares c^2 that are equal to the sums of a number M(n) of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2, starting at b(n) (A002593) for M(n) being twice a squared integer (A001105).

254016, 76055841, 4286582784, 97578140625, 1253342942784, 10849309481241, 70360154505216, 366000785535609, 1599920001000000, 6075789839706609, 20542200096854016, 62996919308080641, 177789795179217984, 467082203214515625, 1152912708530601984

Offset: 2

Author

Vladimir Pletser, Jan 10 2015

Keywords

• nonn
• easy

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 where M(n) is twice a squared integer (A001105) and b(n)=(A002593).

If M is twice a squared integer, there always exists at least one nontrivial solution for the sum of M consecutive cubed integers starting at b^3 and equaling a squared integer c^2. For n>=1, M(n)= 2n^2 (A001105), b(n) = M(M-1)/2 = n^2(2n^2 - 1) (A002593), c(n)= sqrt(M/2) (M(M^2-1)/2)= n^3(4n^4 - 1) (A253724) and a(n)=c(n)^2 (this sequence).

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

Examples

			For n=2, M(n)=8, b(n)=28, c(n)=504, a(n)=c^2=254016.
See "File Triplets (M,b,c) for M=2n^2" link.
		

Links

• Vladimir Pletser, Table of n, a(n) for n = 2..50000
• Vladimir Pletser, File Triplets (M,b,c) for M=2n^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 (15,-105,455,-1365,3003,-5005,6435,-6435,5005,-3003,1365,-455,105,-15,1).

Crossrefs

Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253679, A253680, A253707, A253708, A002593, A253724.

Programs

Magma

[(n^3*(4*n^4-1))^2: n in [2..20]]; // Vincenzo Librandi, Feb 19 2015

Maple

restart: for n from 2 to 50000 do a:=(n^3*(4*n^4 - 1))^2: print (a); end do:

Mathematica

f[n_] := (n^3 (4 n^4 - 1))^2; Rest[f /@ Range@16] (* Michael De Vlieger, Jan 28 2015 *)
LinearRecurrence[{15,-105,455,-1365,3003,-5005,6435,-6435,5005,-3003,1365,-455,105,-15,1},{254016,76055841,4286582784,97578140625,1253342942784,10849309481241,70360154505216,366000785535609,1599920001000000,6075789839706609,20542200096854016,62996919308080641,177789795179217984,467082203214515625,1152912708530601984},20] (* Harvey P. Dale, Feb 18 2023 *)

Formula

a(n) = (n^3(4n^4 - 1))^2.

G.f.: -9*x^2*(x^14 -15*x^13 +106*x^12 +27754*x^11 +8028759*x^10 [phone]*x^9 [phone]*x^8 [phone]*x^7 [phone]*x^6 [phone]*x^5 [phone]*x^4 [phone]*x^3 [phone]*x^2 +8027289*x +28224) / (x -1)^15. - Colin Barker, Jan 14 2015