OFFSET
2,1
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
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).
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)
EXAMPLE
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.
MAPLE
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Russell Walsmith, Apr 26 2009
STATUS
approved