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A257199
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a(n) = n*(n+1)*(n+2)*(n^2+2*n+17)/120.
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4
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1, 5, 16, 41, 91, 182, 336, 582, 957, 1507, 2288, 3367, 4823, 6748, 9248, 12444, 16473, 21489, 27664, 35189, 44275, 55154, 68080, 83330, 101205, 122031, 146160, 173971, 205871, 242296, 283712, 330616, 383537, 443037, 509712, 584193, 667147, 759278, 861328, 974078
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OFFSET
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1,2
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COMMENTS
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Antidiagonal sums of the array of pyramidal numbers shown in Table 2 of Sardelis and Valahas paper (see A261720).
This is the case j = 3 of (n^2 + (j-1)*n + (j+1)^2 + 1)*binomial(n+j-1, j)/((j+1)*(j+2)), where j is the space dimension: a(n) = (n^2+2*n+17)*binomial(n+2,3)/20.
The sequence is the binomial transform of (1, 4, 7, 7, 4, 1, 0, 0, 0, ...). - Gary W. Adamson, Aug 26 2015
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LINKS
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FORMULA
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G.f.: x*(1 - x + x^2)/(1 - x)^6.
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MATHEMATICA
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Table[n (n + 1) (n + 2) (n^2 + 2n + 17)/120, {n, 40}]
LinearRecurrence[{6, -15, 20, -15, 6, -1}, {1, 5, 16, 41, 91, 182}, 40] (* Harvey P. Dale, Mar 18 2018 *)
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PROG
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(Magma) [n*(n+1)*(n+2)*(n^2+2*n+17)/120: n in [1..40]]; // Vincenzo Librandi, Apr 18 2015
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CROSSREFS
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For another version of the array, see A080851.
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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