A174045 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 = 3, read by rows.

1, 1, 1, 1, 6, 1, 1, 24, 24, 1, 1, 70, 230, 70, 1, 1, 90, 881, 881, 90, 1, 1, 231, 2790, 7060, 2790, 231, 1, 1, 295, 8383, 28270, 28270, 8383, 295, 1, 1, 684, 21441, 181680, 242172, 181680, 21441, 684, 1, 1, 750, 58320, 378009, 882549, 882549, 378009, 58320, 750, 1

Offset

1,5

Comments

Row sums are: {1, 2, 8, 50, 372, 1944, 13104, 73898, 649784, 2639258, ...}.

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 = 3.

Example

The triangle begins as:
  1;
  1,   1;
  1,   6,     1;
  1,  24,    24,      1;
  1,  70,   230,     70,      1;
  1,  90,   881,    881,     90,      1;
  1, 231,  2790,   7060,   2790,    231,      1;
  1, 295,  8383,  28270,  28270,   8383,    295,     1;
  1, 684, 21441, 181680, 242172, 181680,  21441,   684,   1;
  1, 750, 58320, 378009, 882549, 882549, 378009, 58320, 750, 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, 3], {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, 3): 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, 3) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Apr 16 2021

Crossrefs

Cf. A174043 (q=1), A174044 (q=2), this sequence (q=3).

Sequence in context: A309280 A155863 A173882 * A169660 A035348 A140945

Adjacent sequences: A174042 A174043 A174044 * A174046 A174047 A174048

Keyword

nonn,tabl,easy,less

Author

Roger L. Bagula, Mar 06 2010

Extensions

Edited by G. C. Greubel, Apr 16 2021

Status

approved