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A112292 An invertible triangle of ratios of double factorials.

%I A112292 #18 Jun 18 2019 08:02:47
%S A112292 1,1,1,3,3,1,15,15,5,1,105,105,35,7,1,945,945,315,63,9,1,10395,10395,
%T A112292 3465,693,99,11,1,135135,135135,45045,9009,1287,143,13,1,2027025,
%U A112292 2027025,675675,135135,19305,2145,195,15,1,34459425,34459425,11486475,2297295,328185,36465,3315,255,17,1
%N A112292 An invertible triangle of ratios of double factorials.
%C A112292 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 A112292 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 A112292 T(n, k)=if(k<=n, (2n-1)!!/(2k-1)!!, 0);
%F A112292 T(n, k)=if(k<=n, n!*C(2n, n)2^(k-n)/(k!*C(2k, k)), 0);
%F A112292 T(n, k)=if(k<=n, 2^(n-k)(n-1/2)!/(k-1/2)!, 0);
%F A112292 T(n, k)=if(k<=n, (n+1)!*C(n)2^(k-n)/((k+1)!*C(k)), 0).
%e A112292 Triangle begins
%e A112292       1;
%e A112292       1,     1;
%e A112292       3,     3,    1;
%e A112292      15,    15,    5,  1;
%e A112292     105,   105,   35,  7,  1;
%e A112292     945,   945,  315, 63,  9,  1;
%e A112292   10395, 10395, 3465,693, 99, 11, 1;
%e A112292 Inverse is A112295, which begins
%e A112292    1;
%e A112292   -1,  1;
%e A112292    0, -3,  1;
%e A112292    0,  0, -5,  1;
%e A112292    0,  0,  0, -7,  1;
%e A112292    0,  0,  0,  0, -9,  1;
%e A112292 Similar results arise for higher factorials.
%t A112292 T[n_, k_] := If[k <= n, (2n-1)!!/(2k-1)!!, 0];
%t A112292 Table[T[n, k], {n, 0, 9}, {k, 0, n}] (* _Jean-François Alcover_, Jun 13 2019 *)
%Y A112292 Columns include A001147, A051577, A051579.
%Y A112292 Row sums are A112293.
%Y A112292 Diagonal sums are A112294.
%Y A112292 Cf. A094587 (p=1), this sequence (p=2), A136214 (p=3).
%K A112292 easy,nonn,tabl
%O A112292 0,4
%A A112292 _Paul Barry_, Sep 01 2005