login
Triangle read by rows: T(n,0) = 1, T(n,k) = 3*binomial(n,k) if k>=0 (0<=k<=n).
3

%I #17 Sep 08 2022 08:45:28

%S 1,1,3,1,6,3,1,9,9,3,1,12,18,12,3,1,15,30,30,15,3,1,18,45,60,45,18,3,

%T 1,21,63,105,105,63,21,3,1,24,84,168,210,168,84,24,3,1,27,108,252,378,

%U 378,252,108,27,3,1,30,135,360,630,756,630,360,135,30,3

%N Triangle read by rows: T(n,0) = 1, T(n,k) = 3*binomial(n,k) if k>=0 (0<=k<=n).

%C Row sums = A033484: (1, 4, 10, 22, 46, 94...); 3*2^n - 2.

%C Analogous triangle using (1,2,2,2...) as the main diagonal of M = A124927.

%C Except for the first column, entries in the Pascal triangle are tripled.

%H G. C. Greubel, <a href="/A124928/b124928.txt">Rows n = 0..100 of triangle, flattened</a>

%F G.f.: G(t,z) = 3/(1-(1+t)*z) - 2/(1-z).

%e First few rows of the triangle are:

%e 1;

%e 1, 3;

%e 1, 6, 3;

%e 1, 9, 9, 3;

%e 1, 12, 18, 12, 3;

%e 1, 15, 30, 30, 15, 3;

%e 1, 18, 45, 60, 45, 18, 3;

%e ...

%p T:=proc(n,k) if k=0 then 1 else 3*binomial(n,k) fi end: for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

%t Flatten[Table[If[k==0,1,3*Binomial[n,k]],{n,0,20},{k,0,n}]] (* _Harvey P. Dale_, Oct 19 2013 *)

%o (PARI) T(n,k) = if(k==0, 1, 3*binomial(n,k)); \\ _G. C. Greubel_, Nov 19 2019

%o (Magma) [k eq 0 select 1 else 3*Binomial(n,k): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Nov 19 2019

%o (Sage)

%o def T(n, k):

%o if (k==0): return 1

%o else: return 3*binomial(n,k)

%o [[T(n, k) for k in (0..n)] for n in (0..12)] # _G. C. Greubel_, Nov 19 2019

%o (GAP)

%o T:= function(n,k)

%o if k=0 then return 1;

%o else return 3*Binomial(n,k);

%o fi; end;

%o Flat(List([0..12], n-> List([0..n], k-> T(n,k) ))); # _G. C. Greubel_, Nov 19 2019

%Y Cf. A033484, A124927.

%K nonn,tabl

%O 0,3

%A _Gary W. Adamson_, Nov 12 2006

%E Edited by _N. J. A. Sloane_, Nov 29 2006