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
Examples
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; ...
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..5150 (rows n = 0..100, flattened)
Programs
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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
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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 *)
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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
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
Extensions
Title changed by G. C. Greubel, Apr 26 2021
Comments