login
An invertible triangle of ratios of double factorials.
6

%I #18 Jun 18 2019 08:02:47

%S 1,1,1,3,3,1,15,15,5,1,105,105,35,7,1,945,945,315,63,9,1,10395,10395,

%T 3465,693,99,11,1,135135,135135,45045,9009,1287,143,13,1,2027025,

%U 2027025,675675,135135,19305,2145,195,15,1,34459425,34459425,11486475,2297295,328185,36465,3315,255,17,1

%N An invertible triangle of ratios of double factorials.

%C As a square array read by antidiagonals, column k has e.g.f. (1/(1-2x)^(1/2))*(1/(1-2x))^k. - _Paul Barry_, Sep 04 2005

%C Let G(m, k, p) = (-p)^k*Product_{j=0..k-1}(j - m - 1/p) and T(n, k, p) = G(n-1, n-k, p) then T(n, k, 1) = A094587(n, k), T(n, k, 2) is this sequence and T(n, k, 3) = A136214. - _Peter Luschny_, Jun 01 2009, revised Jun 18 2019

%F T(n, k)=if(k<=n, (2n-1)!!/(2k-1)!!, 0);

%F T(n, k)=if(k<=n, n!*C(2n, n)2^(k-n)/(k!*C(2k, k)), 0);

%F T(n, k)=if(k<=n, 2^(n-k)(n-1/2)!/(k-1/2)!, 0);

%F T(n, k)=if(k<=n, (n+1)!*C(n)2^(k-n)/((k+1)!*C(k)), 0).

%e Triangle begins

%e 1;

%e 1, 1;

%e 3, 3, 1;

%e 15, 15, 5, 1;

%e 105, 105, 35, 7, 1;

%e 945, 945, 315, 63, 9, 1;

%e 10395, 10395, 3465,693, 99, 11, 1;

%e Inverse is A112295, which begins

%e 1;

%e -1, 1;

%e 0, -3, 1;

%e 0, 0, -5, 1;

%e 0, 0, 0, -7, 1;

%e 0, 0, 0, 0, -9, 1;

%e Similar results arise for higher factorials.

%t T[n_, k_] := If[k <= n, (2n-1)!!/(2k-1)!!, 0];

%t Table[T[n, k], {n, 0, 9}, {k, 0, n}] (* _Jean-François Alcover_, Jun 13 2019 *)

%Y Columns include A001147, A051577, A051579.

%Y Row sums are A112293.

%Y Diagonal sums are A112294.

%Y Cf. A094587 (p=1), this sequence (p=2), A136214 (p=3).

%K easy,nonn,tabl

%O 0,4

%A _Paul Barry_, Sep 01 2005