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Triangle of flag coefficients [ n k ] (denominators of rational parts).
2

%I #16 Oct 22 2020 18:59:16

%S 1,1,1,1,2,1,1,1,1,1,1,4,1,4,1,1,3,1,1,3,1,1,16,1,8,1,16,1,1,5,1,1,1,

%T 1,5,1,1,32,1,32,3,32,1,32,1,1,35,1,5,1,1,5,1,35,1,1,256,1,64,1,128,1,

%U 64,1,256,1,1,63,1,7,3,1,1,3,7,1,63,1,1,512,1,512,1,256,5,256,1,512,1,512

%N Triangle of flag coefficients [ n k ] (denominators of rational parts).

%D D. A. Klain and G.-C. Rota, Introduction to Geometric Probability, Cambridge, p. 69.

%F [ 2m 2k ] = C(2m, 2k)/C(m, k); [ 2m 2k+1 ] = Pi(2m)!/4^m*k!m!(m-k-1)!; [ 2m+1 2k ] = 4^k*C(m, k)/C(2k, k); [ 2m+1 2k+1 ] = 4^(m-k)*C(m, m-k)/C(2m-2k, m-k).

%e Triangle begins

%e 1;

%e 1 1;

%e 1*Pi/2 1;

%e 1 2 2 1;

%e 1 3*Pi/4 3 3*Pi/4 1;

%e 1 8/3 4 4 8/3 1; ...

%t f[m_?EvenQ, k_?EvenQ] := Binomial[m, k]/Binomial[m/2, k/2]; f[m_?EvenQ, k_?OddQ] := (Pi*m!)/(2^m*(((1/2)*(k - 1))!*(-1 + (1 - k)/2 + m/2)!*(m/2)!)); f[m_?OddQ, k_?EvenQ] := (2^k*Binomial[(1/2)*(m - 1), k/2])/ Binomial[k, k/2]; f[m_?OddQ, k_?OddQ] := (4^((1 - k)/2 + (1/2)*(m - 1))* Binomial[(1/2)*(m - 1), (1 - k)/2 + (1/2)*(m - 1)]) / Binomial[m - k , (1 - k)/2 + (1/2)*(m - 1)]; A036065 = Denominator[ Flatten[ Table[f[m, k], {m, 0, 12}, {k, 0, m}] ]](* _Jean-François Alcover_, May 10 2012, from formula *)

%Y Cf. A036064.

%K nonn,easy,tabl,nice,frac

%O 0,5

%A _N. J. A. Sloane_

%E More terms from _Naohiro Nomoto_, Jun 20 2001