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

Author

Russell Walsmith, Apr 26 2009

Keywords

• nonn
• easy

Comments

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

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).

Crossrefs

Cf. A005581, A097900, A159919.

Programs

Magma

[n*(n-1)*(n+4)/3: n in [2..60]]; // G. C. Greubel, Oct 03 2022

Maple

A159920:=n->n*(n-1)*(n+4)/3; seq(A159920(k), k=2..100); # Wesley Ivan Hurt, Oct 19 2013

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