A134059 Triangle T(n, k) = 3*binomial(n,k) with T(0, 0) = 1, read by rows.

1, 3, 3, 3, 6, 3, 3, 9, 9, 3, 3, 12, 18, 12, 3, 3, 15, 30, 30, 15, 3, 3, 18, 45, 60, 45, 18, 3, 3, 21, 63, 105, 105, 63, 21, 3, 3, 24, 84, 168, 210, 168, 84, 24, 3, 3, 27, 108, 252, 378, 378, 252, 108, 27, 3

Offset

0,2

Comments

Triangle T(n,k), 0 <= k <= n, read by rows given by [3, -2, 0, 0, 0, 0, 0, ...] DELTA [3, -2, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 07 2007

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..5150 (rows n = 0..100, flattened)

Formula

3*Pascal's triangle A007318, then replace T(0,0) with 1.

G.f.: Sum_{n>=0} Sum_{k>=0} T(n,k) *x^n * y^k = 1 - 3*(1+y)*x/(x+x*y-1). - R. J. Mathar, Feb 19 2020

From G. C. Greubel, Apr 27 2021: (Start)

T(n, k) = 3*binomial(n,k) - 2*[n=0].

Sum_{k=0..n} T(n, k) = 3*2^n - 2*[n=0] = A082505(n+1). (End)

E.g.f.: 3*exp(x*(1+y)) - 2. - Stefano Spezia, Apr 03 2024

Example

First few rows of the triangle:
  1;
  3,  3;
  3,  6,  3;
  3,  9,  9,  3;
  3, 12, 18, 12,  3;
  3, 15, 30, 30, 15,  3;
  3, 18, 45, 60, 45, 18,  3;
  ...

Mathematica

Join[{1}, Rest[Flatten[Table[3Binomial[n, k], {n, 0, 10}, {k, 0, n}]]]] (* Harvey P. Dale, Feb 15 2014 *)
Table[3*Binomial[n, k] -2*Boole[n==0], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Apr 26 2021 *)

Prog

(Magma)
A134059:= func< n, k | n eq 0 select 1 else 3*Binomial(n, k) >;
[A134059(n, k): k in [0..n], n in [0..12]]; // G. C. Greubel, Apr 26 2021
(Sage)
def A134059(n, k): return 3*binomial(n, k) - 2*bool(n==0)
flatten([[A134059(n, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 26 2021

Crossrefs

Cf. A007318, A082505 (row sums), A084938, A134058.

Sequence in context: A100049 A261926 A158315 * A219915 A219217 A219686

Adjacent sequences: A134056 A134057 A134058 * A134060 A134061 A134062

Keyword

nonn,tabl,easy

Author

Gary W. Adamson, Oct 05 2007

Extensions

Title changed by G. C. Greubel, Apr 26 2021

Status

approved