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%I #15 Oct 12 2020 20:02:36
%S 1,3,5,15,27,33,105,195,261,279,945,1785,2475,2925,2895,10395,19845,
%T 28035,34425,37935,35685,135135,259875,371385,465255,533925,562275,
%U 509985,2027025,3918915,5644485,7158375,8390025,9218475,9401805,8294895,34459425,66891825,96891795,123898005,147093975,165209625,176067675,175313565,151335135,654729075,1274998725,1854727875,2385808425,2857013775,3252014325,3545408475,3693650625,3609649575,3061162125
%N Triangle read by rows: T(n,k) (1 <= k <= n-1, n >= 2) = d(2*(n-k)-1)*(d(2*n-2)/d(2*(n-k)-2) - d(2*n-3)/d(2*(n-k)-3)), where d = A006882 is the double factorial function.
%H N. Ochiumi, <a href="http://infoshako.sk.tsukuba.ac.jp/~hachi/COS/cos2011/abst/ochiumi.pdf">On the total sum of number of nodes covering a given number of leaves in an unordered binary tree</a>.
%e Triangle begins
%e 1,
%e 3, 5,
%e 15, 27, 33,
%e 105, 195, 261, 279,
%e 945, 1785, 2475, 2925, 2895,
%e 10395, 19845, 28035, 34425, 37935, 35685,
%e 135135, 259875, 371385, 465255, 533925, 562275, 509985,
%e ...
%p d:=doublefactorial;
%p a:=(n,k)-> d(2*(n-k)-1)*(d(2*n-2)/d(2*(n-k)-2) - d(2*n-3)/d(2*(n-k)-3));
%p f:=n->[seq(a(n,k),k=1..n-1)];
%p for n from 1 to 10 do lprint(f(n)); od:
%t a[n_, k_] := (2*(n-k)-1)!!*((2*n-2)!!/(2*(n-k)-2)!!-(2*n-3)!!/(2*(n-k)-3)!!); Table[a[n, k], {n, 2, 11}, {k, 1, n-1}] // Flatten (* _Jean-François Alcover_, Jan 08 2014 *)
%Y Edges of triangle are A006882 and A129890.
%K nonn,tabl
%O 2,2
%A _N. J. A. Sloane_, Dec 14 2011
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