A174447 Triangle T(n, k, q) = ceiling(binomial(n, k)/f(n, q)) with T(0, 0) = 1, f(n, q) = 1 + tanh((n-1)/q), and q = 4, read by rows.

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 3, 4, 3, 1, 1, 3, 6, 6, 3, 1, 1, 4, 9, 11, 9, 4, 1, 1, 4, 12, 19, 19, 12, 4, 1, 1, 5, 15, 29, 37, 29, 15, 5, 1, 1, 5, 19, 43, 65, 65, 43, 19, 5, 1, 1, 6, 23, 61, 107, 128, 107, 61, 23, 6, 1, 1, 6, 28, 84, 167, 233, 233, 167, 84, 28, 6, 1

Offset

0,5

Links

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Formula

T(n, k, q) = ceiling(binomial(n, k)/f(n, q)) with T(0, 0) = 1, f(n, q) = 1 + tanh((n-1)/q), and q = 4.

Example

Triangle begins as:
  1;
  1, 1;
  1, 2,  1;
  1, 3,  3,  1;
  1, 3,  4,  3,   1;
  1, 3,  6,  6,   3,   1;
  1, 4,  9, 11,   9,   4,   1;
  1, 4, 12, 19,  19,  12,   4,  1;
  1, 5, 15, 29,  37,  29,  15,  5,  1;
  1, 5, 19, 43,  65,  65,  43, 19,  5, 1;
  1, 6, 23, 61, 107, 128, 107, 61, 23, 6, 1;

Mathematica

f[n_, q_]= 1 + Tanh[(n-1)/q];
T[n_, k_, q_]= If[n==0, 1, Ceiling[Binomial[n, k]/f[n, q]]];
Table[T[n, k, 4], {n, 0, 12}, {k, 0, n}]//Flatten

Prog

(Magma)
T:= func< n, k, q | n eq 0 select 1 else Ceiling(Binomial(n, k)/(1 + Tanh((n-1)/q))) >;
[T(n, k, 4): k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 05 2021
(Sage)
def T(n, k, q): return 1 if (n==0) else ceil(binomial(n, k)/(1 + tanh((n-1)/q)))
flatten([[T(n, k, 4) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Aug 05 2021

Crossrefs

Cf. A174446 (q=1), this sequence (q=4), A174448 (q=12).

Sequence in context: A160832 A351522 A157458 * A174374 A242641 A347187

Adjacent sequences: A174444 A174445 A174446 * A174448 A174449 A174450

Keyword

nonn,tabl

Author

Roger L. Bagula, Mar 20 2010

Extensions

Edited by G. C. Greubel, Aug 05 2021

Status

approved