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A253708
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
323, 7497, 57618, 262430, 878445, 2399103, 5669972, 12026988, 23457735, 42785765, 73877958, 121874922, 193444433, 297057915, 443289960, 645140888, 918382347, 1281925953, 1758214970, 2373639030, 3158971893, 4149832247, 5387167548, 6917760900, 8794760975
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 b being an odd squared integer (A016754).
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).
The trivial solutions with M < 1 and b < 2 are not considered here.
LINKS
FORMULA
a(n) = (n(n+1)/2)(4(2n+1)^4-1).
G.f.: -x*(323*x^4+5236*x^3+11922*x^2+5236*x+323) / (x-1)^7. - Colin Barker, Jan 14 2015
EXAMPLE
For n=1, b(n)=9, M(n)=17, a(n)=323.
See File Triplets (M,b,c) for a=(2n+1)^2</a>" link.
MAPLE
restart: for n from 1 to 50000 do a:= (n*(n+1)/2)(4*(2*n+1)^4-1): print (a); end do:
MATHEMATICA
f[n_] := (n (n + 1)/2) (4 (2 n + 1)^4 - 1); Array[f, 33] (* Michael De Vlieger, Jan 10 2015 *)
PROG
(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
(Magma) [(n*(n+1)/2)*(4*(2*n+1)^4-1): n in [1..30]]; // Vincenzo Librandi, Feb 19 2015
KEYWORD
nonn,easy
AUTHOR
Vladimir Pletser, Jan 09 2015
STATUS
approved