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%I #8 Apr 24 2014 14:19:48
%S 1,1,2,1,1,3,1,2,4,8,1,1,3,3,45,1,2,12,8,144,288,1,1,1,1,10,10,140,1,
%T 2,12,24,720,160,8640,17280,1,1,3,1,45,45,945,189,14175,1,2,4,8,80,32,
%U 2240,4480,6400,89600,1,1,3,3,18,2,756,189,4536,4536,299376
%N Triangle read by rows: T(n,k) (0 <= k <= n) = denominator of Integral_{x=0..n} binomial(x,k).
%C Suggested by the integral formula for the Cotesian numbers A100640/A100641.
%e Triangle of fractions Integral_{x=0..n} binomial(x,k) begins:
%e [0],
%e [1, 1/2],
%e [2, 2, 1/3],
%e [3, 9/2, 9/4, 3/8],
%e [4, 8, 20/3, 8/3, 14/45],
%e [5, 25/2, 175/12, 75/8, 425/144, 95/288],
%e [6, 18, 27, 24, 123/10, 33/10, 41/140],
%e [7, 49/2, 539/12, 1225/24, 26117/720, 2499/160, 30919/8640, 5257/17280],
%e [8, 32, 208/3, 96, 3928/45, 2336/45, 18128/945, 736/189, 3956/14175],
%e ...
%p T:=proc(n,k) integrate( expand(binomial(x,k)), x=0..n); end;
%p t0:=[seq( [seq(T(n,k),k=0..n)], n=0..10)];
%Y Cf. A241186, A100640, A100641.
%K nonn,tabl,frac
%O 0,3
%A _N. J. A. Sloane_, Apr 24 2014
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