|
%I #33 Jan 07 2019 04:27:16
%S 1,1,1,1,4,1,1,6,6,1,1,8,12,8,1,1,10,20,20,10,1,1,12,30,40,30,12,1,1,
%T 14,42,70,70,42,14,1,1,16,56,112,140,112,56,16,1,1,18,72,168,252,252,
%U 168,72,18,1,1,20,90,240,420,504,420,240,90,20,1
%N Triangle read by rows: T(n,0) = T(n,n) = 1, and T(n,k) = 2*binomial(n,k) for 1 <= k <= n - 1.
%C T(2*n,n) is A100320 (with Hankel transform A144704). - _Paul Barry_, Sep 19 2008
%C Double the internal elements of Pascal's triangle. - _Paul Barry_, Jan 07 2009
%C Coefficients of 2*(x + 1)^n - (x^n + 1) as a triangle (except for the very first term). - _Thomas Baruchel_, Jun 02 2018
%F T(n,k) = 2*A007318(n,k) - A103451(n,k).
%F T(n,k) = [k<=n] (0^(n + k) + C(n,k)*(2 - 0^(n - k) - 0^k)). - _Paul Barry_, Sep 19 2008
%F T(n,k) = A007318(n,k)*A154325(n,k). - _Paul Barry_, Jan 07 2009
%F From _Emanuele Munarini_, May 15 2018: (Start)
%F G.f.: (1 - t - x*t + 3*x*t^2 - x*t^3 - x^2*t^3)/((1 - t)*(1 - x*t)*(1 - t - x*t)).
%F T(n+3,k+2) = 2*T(n+2,k+2) - T(n+1,k+2) + 2*T(n+2,k+1) - 3*T(n+1,k+1) - T(n+1,k) + T(n,k+1) + T(n,k), except for n = 0 and k = 0. (End)
%F E.g.f.: 1 - exp(t) - exp(t*x) + 2*exp(t*(1 + x)). - _Franck Maminirina Ramaharo_, Jan 02 2019
%e First few rows of the triangle are:
%e 1;
%e 1, 1;
%e 1, 4, 1;
%e 1, 6, 6, 1;
%e 1, 8, 12, 8, 1;
%e 1, 10, 20, 20, 10, 1;
%e 1, 12, 30, 40, 30, 12, 1;
%e 1, 14, 42, 70, 70, 42, 14, 1;
%e ...
%t T[n_, k_] := If[n == k || k == 0, 1, If[k <= n, 2 Binomial[n, k], 0]]
%t Flatten[Table[T[n, k], {n, 0, 20}, {k, 0, n}]] (* _Emanuele Munarini_, May 15 2018 *)
%o (Maxima) T(n, k) := if k = 0 or k = n then 1 else 2*binomial(n, k)$
%o create_list(T(n, k), n, 0, 20, k, 0, n); /* _Franck Maminirina Ramaharo_, Jan 03 2019 */
%Y Row sums: A095121.
%Y Cf. A007318, A103451, A132046.
%Y Cf. A154327 (diagonal sums). [_Paul Barry_, Jan 07 2009]
%Y Cf. A141540.
%K nonn,easy,tabl
%O 0,5
%A _Gary W. Adamson_, Aug 08 2007
|
