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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).
7
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
1,1
COMMENTS
Numbers c(n) such that b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2 has nontrivial solutions over the integers for M being an odd positive integer.
To every odd positive integer M corresponds a sum of M consecutive cubed integers starting at b^3 having at least one nontrivial solution. For n>=1, M(n)=(2n+1) (A005408), b(n) = M^3 -(3M-1)/2 = (2n+1)^3 - (3n+1) (A253679) 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 b < 2 are not considered here.
Stroeker stated that all odd values of M yield a solution to b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2. This was further demonstrated by Pletser.
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
EXAMPLE
For n=1, M(n)=3, b(n)=23, c(n)=204.
See "File Triplets (M,b,c) for M=(2n+1)" link.
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:
MATHEMATICA
f[n_] := 2 n (n + 1) (2 n + 1) (8 n (n + 1) + 1); Array[f, 36] (* Michael De Vlieger, Jan 10 2015 *)
PROG
(PARI) Vec(12*x*(x+1)*(17*x^2+126*x+17)/(x-1)^6 + O(x^100)) \\ Colin Barker, Jan 09 2015
(Magma) [2*n*(n+1)*(2*n+1)*(8*n*(n+1)+1): n in [1..30]]; // Vincenzo Librandi, Feb 19 2015
KEYWORD
nonn,easy
AUTHOR
Vladimir Pletser, Jan 08 2015
STATUS
approved