A159920 Sums of the antidiagonals of Sundaram's sieve (A159919).
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
Offset: 2
Examples
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.
Links
- G. C. Greubel, Table of n, a(n) for n = 2..1000
- Andrew Baxter, Sundaram's Sieve.
- Julian Havil, Sundaram's Sieve, Plus Magazine, March 2009.
- New Zealand Maths, Newletter 18, October 2002.
- Wikipedia, Sundaram's Sieve.
- Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Programs
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Magma
[n*(n-1)*(n+4)/3: n in [2..60]]; // G. C. Greubel, Oct 03 2022
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Maple
A159920:=n->n*(n-1)*(n+4)/3; seq(A159920(k), k=2..100); # Wesley Ivan Hurt, Oct 19 2013
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Mathematica
Table[(n-1)*n*(n+4)/3,{n,2,60}] (* Vladimir Joseph Stephan Orlovsky, Apr 28 2010 *) LinearRecurrence[{4,-6,4,-1},{4,14,32,60},61] (* Harvey P. Dale, Apr 23 2011 *) -
SageMath
[n*(n-1)*(n+4)/3 for n in range(2,60)] # G. C. Greubel, Oct 03 2022
Formula
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).
a(n) = 2*A005581(n), n > 1.
a(n) = Sum_{i=1..n-1} i*(i + 3). - Wesley Ivan Hurt, Oct 19 2013
From G. C. Greubel, Oct 03 2022: (Start)
G.f.: 2*x^2*(2 - x)/(1-x)^4.
E.g.f.: (1/3)*x^2*(6 + x)*exp(x). (End)
a(n) = 2*A097900(n)/(n-2)! for n >= 2. - Cullen M. Vaney, Jul 14 2025
Comments