A174043 Triangle T(n, k, q) = Sum_{j=0..10} q^j * floor( binomial(n+1,k)*binomial(n-1,k-1)/(2^j*(n+1)) ) for q = 1, read by rows.

1, 1, 1, 1, 4, 1, 1, 10, 10, 1, 1, 18, 38, 18, 1, 1, 26, 97, 97, 26, 1, 1, 39, 206, 344, 206, 39, 1, 1, 53, 389, 974, 974, 389, 53, 1, 1, 70, 669, 2348, 3522, 2348, 669, 70, 1, 1, 86, 1076, 5033, 10575, 10575, 5033, 1076, 86, 1, 1, 105, 1644, 9890, 27704, 38784, 27704, 9890, 1644, 105, 1

Offset

1,5

Comments

Row sums are: {1, 2, 6, 22, 76, 248, 836, 2834, 9698, 33542, ...}.

Links

G. C. Greubel, Rows n = 1..50 of the triangle, flattened

Formula

T(n, k, q) = Sum_{j=0..10} q^j * floor( binomial(n+1,k)*binomial(n-1,k-1)/(2^j*(n+1)) ) for q = 1.

Example

Triangle begins as:
  1;
  1,  1;
  1,  4,    1;
  1, 10,   10,    1;
  1, 18,   38,   18,     1;
  1, 26,   97,   97,    26,     1;
  1, 39,  206,  344,   206,    39,    1;
  1, 53,  389,  974,   974,   389,   53,    1;
  1, 70,  669, 2348,  3522,  2348,  669,   70,  1;
  1, 86, 1076, 5033, 10575, 10575, 5033, 1076, 86, 1;

Mathematica

T[n_, k_, q_]:= Sum[q^j*Floor[Binomial[n-1, k-1]*Binomial[n+1, k]/(2^j*(n+1))], {j, 0, 10}];
Table[T[n, k, 1], {n, 12}, {k, n}]//Flatten (* modified by G. C. Greubel, Apr 16 2021 *)

Prog

(Magma)
T:= func< n, k, q | (&+[ q^j*Floor( Binomial(n+1, k)*Binomial(n-1, k-1)/(2^j*(n+1)) ): j in [0..10]]) >;
[T(n, k, 1): k in [1..n], n in [1..12]]; // G. C. Greubel, Apr 16 2021
(Sage)
def T(n, k, q): return sum( q^j*( (binomial(n+1, k)*binomial(n-1, k-1)//(2^j*(n+1))) ) for j in (0..10))
flatten([[T(n, k, 1) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Apr 16 2021

Crossrefs

Cf. this sequence (q=1), A174044 (q=2), A174045 (q=3).

Sequence in context: A109955 A214398 A220860 * A319029 A175124 A089447

Adjacent sequences: A174040 A174041 A174042 * A174044 A174045 A174046

Keyword

nonn,tabl,easy,less

Author

Roger L. Bagula, Mar 06 2010

Extensions

Edited by G. C. Greubel, Apr 16 2021

Status

approved