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A175831
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Partial sums of ceiling(n^2/12).
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1
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0, 1, 2, 3, 5, 8, 11, 16, 22, 29, 38, 49, 61, 76, 93, 112, 134, 159, 186, 217, 251, 288, 329, 374, 422, 475, 532, 593, 659, 730, 805, 886, 972, 1063, 1160, 1263, 1371, 1486, 1607, 1734, 1868, 2009, 2156, 2311, 2473, 2642, 2819, 3004, 3196, 3397, 3606
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OFFSET
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0,3
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COMMENTS
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There are several sequences of integers of the form ceiling(n^2/k) for whose partial sums we can establish identities as following (only for k = 2,...,8,10,11,12, 14,15,16,19,20,23,24).
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LINKS
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FORMULA
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a(n) = round((2*n+1)*(2*n^2 + 2*n + 41)/144).
a(n) = floor((n+1)*(2*n^2 + n + 41)/72).
a(n) = ceiling((2*n^3 + 3*n^2 + 42*n)/72).
a(n) = a(n-12) + (n+1)*(n-12) + 61.
G.f.: x*(1-x^2+x^4) / ( (1+x)*(1+x+x^2)*(x-1)^4 ). - R. J. Mathar, Jun 22 2011
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EXAMPLE
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a(12) = 0 + 1 + 1 + 1 + 2 + 3 + 3 + 5 + 6 + 7 + 9 + 11 + 12 = 61.
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MAPLE
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seq(floor((n+1)*(2*n^2+n+41)/72), n=0..50)
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MATHEMATICA
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Accumulate[Ceiling[Range[0, 50]^2/12]] (* or *) LinearRecurrence[{2, 0, -1, -1, 0, 2, -1}, {0, 1, 2, 3, 5, 8, 11}, 60] (* Harvey P. Dale, Apr 16 2023 *)
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PROG
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(Magma) [Round((2*n+1)*(2*n^2+2*n+41)/144): n in [0..60]]; // Vincenzo Librandi, Jun 22 2011
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CROSSREFS
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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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