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

0,2

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

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

Mathematica

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

Prog

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

Crossrefs

Cf. A065942, A081181, A081205.

Sequence in context: A369781 A026336 A027913 * A293308 A357269 A106672

Adjacent sequences: A081201 A081202 A081203 * A081205 A081206 A081207

Keyword

easy,nonn

Author

Paul Barry, Mar 11 2003

Status

approved