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A159920
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Sums of the antidiagonals of Sundaram's sieve (A159919).
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6
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4, 14, 32, 60, 100, 154, 224, 312, 420, 550, 704, 884, 1092, 1330, 1600, 1904, 2244, 2622, 3040, 3500, 4004, 4554, 5152, 5800, 6500, 7254, 8064, 8932, 9860, 10850, 11904, 13024, 14212, 15470, 16800, 18204, 19684, 21242, 22880, 24600, 26404
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OFFSET
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2,1
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COMMENTS
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For every n >= 2, a(n) is the sum of numbers in the (n-1)-th antidiagonal of the Sundaram sieve. (It is not clear why the offset was set to 2 rather than 1.) Thus, if T(j, k) is the element in row j and column k of the Sundaram sieve, we have a(n) = Sum_{i = 1..n-1} T(i, n-i) = Sum_{i = 1..n-1} (2*i*(n-i) + i + (n-i)) = (n - 1)*n*(n + 4)/3 for the sum of the numbers in the (n-1)-th antidiagonal. - Petros Hadjicostas, Jun 19 2019
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LINKS
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FORMULA
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a(n) = (n - 1)*n*(n + 4)/3.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
G.f.: 2*x^2*(2 - x)/(1-x)^4.
E.g.f.: (1/3)*x^2*(6 + x)*exp(x). (End)
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EXAMPLE
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For n = 5, (4*5*9)/3 = 60. Indeed, T(1, 4) + T(2, 3) + T(3, 2) + T(4, 1) = 13 + 17 + 17 + 13 = 60 for the sum of the terms in the 4th antidiagonal of the Sundaram sieve.
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MAPLE
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MATHEMATICA
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LinearRecurrence[{4, -6, 4, -1}, {4, 14, 32, 60}, 61] (* Harvey P. Dale, Apr 23 2011 *)
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PROG
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(Magma) [n*(n-1)*(n+4)/3: n in [2..60]]; // G. C. Greubel, Oct 03 2022
(SageMath) [n*(n-1)*(n+4)/3 for n in range(2, 60)] # G. C. Greubel, Oct 03 2022
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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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