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A120328
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Sum of three consecutive squares: a(n) = n^2 + (n + 1)^2 + (n + 2)^2.
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14
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2, 5, 14, 29, 50, 77, 110, 149, 194, 245, 302, 365, 434, 509, 590, 677, 770, 869, 974, 1085, 1202, 1325, 1454, 1589, 1730, 1877, 2030, 2189, 2354, 2525, 2702, 2885, 3074, 3269, 3470, 3677, 3890, 4109, 4334, 4565, 4802, 5045, 5294, 5549, 5810, 6077, 6350
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OFFSET
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-1,1
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COMMENTS
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A rectangular prism with sides n, n + 1, and n + 2 will have four diagonals of different lengths. The sum of the squares of all four is three times the numbers in this sequence beginning with 14 (third term in the sequence for n = 1). - J. M. Bergot, Sep 15 2011
This sequence differs from A005918 only in the first term.
a(n) is also defined for any negative number and a(-n) = a(n-2).
If a 2-set Y and a 3-set Z are disjoint subsets of an n-set (n >= 5) X then a(n-5) is the number of 4-subsets of X intersecting both Y and Z (from comment in A005918 by Milan Janjic, Sep 08 2007).
(End)
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LINKS
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FORMULA
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O.g.f.: (2 - x + 5*x^2)/(x*(1 - x)^3). (End)
a(n) = 3*(n + 1)^2 + 2 == 2 (mod 3), hence a(n) is never square.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for all n in Z. (End)
Sum_{n>=-1} 1/a(n) = coth(sqrt(2/3)*Pi)*Pi/(2*sqrt(6)) + 1/4.
Sum_{n>=-1} (-1)^(n+1)/a(n) = cosech(sqrt(2/3)*Pi)*Pi/(2*sqrt(6)) + 1/4. (End)
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MAPLE
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[seq(n^2+(n+1)^2+(n+2)^2, n=-1..45)];
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MATHEMATICA
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Table[Total[Range[n, n + 2]^2], {n, -1, 45}] (* Harvey P. Dale, Jan 23 2011 *)
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PROG
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(Sage) [i^2+(i+1)^2+(i+2)^2 for i in range(-1, 46)] # Zerinvary Lajos, Jul 03 2008
(PARI) a(n) = n^2 + (n + 1)^2 + (n + 2)^2; \\ Altug Alkan, Nov 11 2015
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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