A253680 Numbers c(n) whose square are equal to the sum of an odd number M of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c(n)^2, starting at b(n) (A253679).
204, 2940, 16296, 57960, 159060, 368004, 754320, 1412496, 2465820, 4070220, 6418104, 9742200, 14319396, 20474580, 28584480, 39081504, 52457580, 69267996, 90135240, 115752840, 146889204, 184391460, 229189296, 282298800, 344826300, 417972204, 503034840
Offset: 1
Examples
For n=1, M(n)=3, b(n)=23, c(n)=204. See "File Triplets (M,b,c) for M=(2n+1)" link.
Links
- Vladimir Pletser, Table of n, a(n) for n = 1..50000
- Vladimir Pletser, File Triplets (M,b,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 (6,-15,20,-15,6,-1).
Programs
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Magma
[2*n*(n+1)*(2*n+1)*(8*n*(n+1)+1): n in [1..30]]; // Vincenzo Librandi, Feb 19 2015
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Maple
restart: for n from 1 to 50000 do c:=2*n*(n+1)*(2*n+1)*(8*n*(n+1)+1): print (c); end do:
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Mathematica
f[n_] := 2 n (n + 1) (2 n + 1) (8 n (n + 1) + 1); Array[f, 36] (* Michael De Vlieger, Jan 10 2015 *)
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PARI
Vec(12*x*(x+1)*(17*x^2+126*x+17)/(x-1)^6 + O(x^100)) \\ Colin Barker, Jan 09 2015
Formula
c(n) = 2n(n+1)*(2n+1)*(8n*(n+1)+1).
G.f.: 12*x*(x+1)*(17*x^2+126*x+17) / (x-1)^6. - Colin Barker, Jan 09 2015
Comments