A253681 Integer squares c^2 that are equal to the sum of an odd number M of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2 starting at b(n) (A253679).
41616, 8643600, 265559616, 3359361600, 25300083600, 135426944016, 568998662400, 1995144950016, 6080268272400, 16566690848400, 41192058954816, 94910460840000, 205045101804816, 419208426176400, 817072496870400, 1527363954902016, 2751797699456400, 4798055269856016
Offset: 1
Examples
For n=1, M(1)=3, b(1)=23, c(1)=204, a(1)=c^2=41616. See "File Triplets (M,b,c) for M=(2n+1)" link, [where in this File, M is the number of term, a the first term and c the square root of the sum].
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, 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 (11,-55,165,-330,462,-462,330,-165,55,-11,1).
Programs
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Magma
[(2*n*(n+1)*(2*n+1)*(8*n*(n+1)+1))^2: n in [1..20]]; // Vincenzo Librandi, Feb 19 2015
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Maple
restart: for n from 1 to 50000 do a:=(2*n*(n+1)*(2*n+1)*(8*n*(n+1)+1))^2: print (a); end do:
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Mathematica
f[n_] := (2 n (n + 1) (2 n + 1) (8 n (n + 1) + 1))^2; Array[f, 21] (* Michael De Vlieger, Jan 10 2015 *)
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PARI
Vec(-144*x*(289*x^8 +56846*x^7 +1199784*x^6 +6296786*x^5 +10697390*x^4 +6296786*x^3 +1199784*x^2 +56846*x +289) / (x -1)^11 + O(x^100)) \\ Colin Barker, Jan 09 2015
Formula
a(n) = (2n(n+1)*(2n+1)*(8n*(n+1)+1))^2.
G.f.: -144*x*(289*x^8 + 56846*x^7 + 1199784*x^6 + 6296786*x^5 + 10697390*x^4 + 6296786*x^3 + 1199784*x^2 + 56846*x + 289) / (x -1)^11. - Colin Barker, Jan 09 2015
Comments