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An invertible triangle of ratios of triple factorials.
6

%I #16 Dec 09 2016 06:19:11

%S 1,2,1,10,5,1,80,40,8,1,880,440,88,11,1,12320,6160,1232,154,14,1,

%T 209440,104720,20944,2618,238,17,1,4188800,2094400,418880,52360,4760,

%U 340,20,1,96342400,48171200,9634240,1204280,109480,7820,460,23,1,2504902400

%N An invertible triangle of ratios of triple factorials.

%C First column is A008544. Second column is A034000. Third column is A051605. As a square array read by antidiagonals, columns have e.g.f. (1/(1-3x)^(2/3)) * (1/(1-3x))^k.

%F Number triangle T(n, k)=if(k<=n, Product{k=1..n, 3k-1}/Product{j=1..k, 3j-1}, 0); T(n, k)=if(k<=n, 3^(n-k)*(n-1/3)!/(k-1/3)!, 0).

%e Triangle begins

%e 1;

%e 2, 1;

%e 10, 5, 1;

%e 80, 40, 8, 1;

%e 880, 440, 88, 11, 1;

%e 12320, 6160, 1232, 154, 14, 1;

%e Inverse triangle A112334 begins

%e 1;

%e -2, 1;

%e 0, -5, 1;

%e 0, 0, -8, 1;

%e 0, 0, 0, -11, 1;

%e 0, 0, 0, 0, -14, 1;

%e 0, 0, 0, 0, 0, -17, 1;

%p nmax:=8: for n from 0 to nmax do for k from 0 to n do if k<=n then T(n, k) := mul(3*k1-1, k1=1..n)/ mul(3*j-1, j=1..k) else T(n, k) := 0: fi: od: od: for n from 0 to nmax do seq(T(n, k), k=0..n) od: seq(seq(T(n, k), k=0..n), n=0..nmax); # _Johannes W. Meijer_, Jul 04 2011, revised Nov 23 2012

%K easy,nonn,tabl

%O 0,2

%A _Paul Barry_, Sep 04 2005