%I #26 Jan 11 2023 11:09:07
%S 1,1,1,3,1,1,10,3,1,1,35,10,3,1,1,126,35,10,3,1,1,462,126,35,10,3,1,1,
%T 1716,462,126,35,10,3,1,1,6435,1716,462,126,35,10,3,1,1,24310,6435,
%U 1716,462,126,35,10,3,1,1,92378,24310,6435,1716,462,126,35,10,3,1,1
%N Number triangle T(n,k) = (-1)^(n-k)*binomial(k-n, n-k) = (0^(n-k) + binomial(2*(n-k), n-k))/2 if k <= n, 0 otherwise; Riordan array (1/(2-C(x)), x) where C(x) is g.f. for Catalan numbers A000108.
%C Triangle includes A088218.
%C Inverse is A106270.
%H G. C. Greubel, <a href="/A106268/b106268.txt">Rows n = 0..50 of the triangle, flattened</a>
%F T(n, k) = (-1)^(n-k)*binomial(k-n, n-k).
%F T(n, k) = (1/2)*(0^(n-k) + binomial(2*(n-k), n-k)).
%F Sum_{k=0..n} T(n, k) = A024718(n) (row sums).
%F Sum_{k=0..floor(n/2)} T(n-k, k) = A106269(n) (diagonal sums).
%F Bivariate g.f.: Sum_{n, k >= 0} T(n,k)*x^n*y^k = (1/2) * (1/(1 - x*y)) * (1 + 1/sqrt(1 - 4*x)). - _Petros Hadjicostas_, Jul 15 2019
%e Triangle (with rows n >= 0 and columns k >= 0) begins as follows:
%e 1;
%e 1, 1;
%e 3, 1, 1;
%e 10, 3, 1, 1;
%e 35, 10, 3, 1, 1;
%e 126, 35, 10, 3, 1, 1;
%e ...
%e Production matrix begins:
%e 1, 1;
%e 2, 0, 1;
%e 5, 0, 0, 1;
%e 14, 0, 0, 0, 1;
%e 42, 0, 0, 0, 0, 1;
%e 132, 0, 0, 0, 0, 0, 1;
%e 429, 0, 0, 0, 0, 0, 0, 1;
%e ... - _Philippe Deléham_, Oct 02 2014
%t T[n_, k_]:= (-1)^(n-k)*Binomial[k-n, n-k];
%t Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, Jan 10 2023 *)
%o (PARI) trg(nn) = {for (n=1, nn, for (k=1, n, print1(binomial(k-n,n-k)*(-1)^(n-k), ", ");); print(););} \\ _Michel Marcus_, Oct 03 2014
%o (Magma)
%o A106268:= func< n,k | k eq n select 1 else (n-k+1)*Catalan(n-k)/2 >;
%o [A106268(n,k): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Jan 10 2023
%o (SageMath)
%o def A106268(n,k): return (1/2)*(0^(n-k) + (n-k+1)*catalan_number(n-k))
%o flatten([[A106268(n,k) for k in range(n+1)] for n in range(13)]) # _G. C. Greubel_, Jan 10 2023
%Y Cf. A000108, A024718 (row sums), A088218, A106269 (diagonal sums), A106270.
%K easy,nonn,tabl
%O 0,4
%A _Paul Barry_, Apr 28 2005