A081204 Staircase on Pascal's triangle.

1, 2, 3, 10, 15, 56, 84, 330, 495, 2002, 3003, 12376, 18564, 77520, 116280, 490314, 735471, 3124550, 4686825, 20030010, 30045015, 129024480, 193536720, 834451800, 1251677700, 5414950296, 8122425444, 35240152720, 52860229080, 229911617056

Offset: 0

Author

Paul Barry, Mar 11 2003

Keywords

• easy
• nonn

Comments

Arrange Pascal's triangle as a square array. This sequence is then a diagonal staircase on the square array.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200

Crossrefs

Cf. A065942, A081181, A081205.

Programs

Magma

[Binomial(Ceiling((n)/2) + n, n): n in [0..30]]; // Vincenzo Librandi, Aug 07 2013

Mathematica

Table[Binomial[Ceiling[(n)/2] + n, n], {n, 0, 30}] (* Vincenzo Librandi, Aug 07 2013 *)

Formula

a(n) = binomial(ceiling((n)/2) + n, n).

a(n) = Sum_{k=0..floor(n/2)} binomial(n+k-1, k). - Paul Barry, Jul 06 2004

Conjecture: 4*n*(n+1)*(6*n^2 - 15*n + 8)*a(n) + 6*n*(9*n-7)*a(n-1) - 3*(3*n-4)*(3*n-2)*(6*n^2-3*n-1)*a(n-2) = 0. - R. J. Mathar, Nov 07 2014

Conjecture: 8*n^2*(n+1)*a(n) - 12*n*(83*n^2 - 313*n + 232)*a(n-1) + 6*(-9*n^3 - 377*n + 384)*a(n-2) + 9*(3*n-5)*(83*n-64)*(3*n-7)*a(n-3) = 0. - R. J. Mathar, Nov 07 2014