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Numbers c(n) whose squares are equal to the sums of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2, for a first term b(n) being an odd squared integer (A016754).
5

%I #43 Jan 08 2025 10:58:26

%S 323,7497,57618,262430,878445,2399103,5669972,12026988,23457735,

%T 42785765,73877958,121874922,193444433,297057915,443289960,645140888,

%U 918382347,1281925953,1758214970,2373639030,3158971893,4149832247,5387167548,6917760900,8794760975

%N Numbers c(n) whose squares are equal to the sums of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2, for a first term b(n) being an odd squared integer (A016754).

%C Numbers c(n) 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).

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

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

%H Vladimir Pletser, <a href="/A253708/b253708.txt">Table of n, a(n) for n = 1..50000</a>

%H Vladimir Pletser, <a href="/A253707/a253707_1.txt">File Triplets (M,b,c) for a=(2n+1)^2</a>

%H Vladimir Pletser, <a href="http://www.researchgate.net/profile/Vladimir_Pletser/publication/271272786">Number of terms, first term and square root of sums of consecutive cubed integers equal to integer squares</a>, Research Gate, 2015.

%H Vladimir Pletser, <a href="http://arxiv.org/abs/1501.06098">General solutions of sums of consecutive cubed integers equal to squared integers</a>, arXiv:1501.06098 [math.NT], 2015.

%H R. J. Stroeker, <a href="http://www.numdam.org/item?id=CM_1995__97_1-2_295_0">On the sum of consecutive cubes being a perfect square</a>, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (7,-21,35,-35,21,-7,1).

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

%F G.f.: -x*(323*x^4+5236*x^3+11922*x^2+5236*x+323) / (x-1)^7. - _Colin Barker_, Jan 14 2015

%e For n=1, b(n)=9, M(n)=17, a(n)=323.

%e See "File Triplets (M,b,c) for a=(2n+1)^2" link.

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

%t f[n_] := (n (n + 1)/2) (4 (2 n + 1)^4 - 1); Array[f, 33] (* _Michael De Vlieger_, Jan 10 2015 *)

%o (PARI) Vec(-x*(323*x^4+5236*x^3+11922*x^2+5236*x+323)/(x-1)^7 + O(x^100)) \\ _Colin Barker_, Jan 14 2015

%o (Magma) [(n*(n+1)/2)*(4*(2*n+1)^4-1): n in [1..30]]; // _Vincenzo Librandi_, Feb 19 2015

%Y Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253679, A253681, A253707, A253709.

%K nonn,easy,changed

%O 1,1

%A _Vladimir Pletser_, Jan 09 2015