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b[n_] := b[n] = Expand[If[n == 0, 1, Sum[b[n-j]*Binomial[n-1, j-1]*If[ OddQ[j], (j-1)!, x+(j-1)!-1], {j, 1, n}]]];
T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][ b[n]];
Table[T[n], {n, 0, 14}] // Flatten (* Jean-François Alcover, May 03 2017, after Alois P. Heinz *)
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Alois P. Heinz, <a href="/A186764/b186764.txt">Rows n = 0..200, flattened</a>
E.g.f.: G(t,z) = exp((t-1)(cosh z - 1))/(1-z).
H(x,y,u,v,z) = exp(((x-u)sinh z + (y-v)(cosh z - 1))*(1+z)^{(u-v)/2}/(1-z)^{(u+v)/2}.
We have: G(t,z) = H(1,t,1,1,z).
# second Maple program:
b:= proc(n) option remember; expand(
`if`(n=0, 1, add(b(n-j)*binomial(n-1, j-1)*
`if`(j::odd, (j-1)!, x+((j-1)!-1)), j=1..n)))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n)):
seq(T(n), n=0..14); # Alois P. Heinz, Apr 13 2017
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_Emeric Deutsch (deutsch(AT)duke.poly.edu), _, Feb 27 2011
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allocated for Emeric DeutschTriangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} having k increasing even cycles (0<=k<=n). A cycle (b(1), b(2), ...) is said to be increasing if, when written with its smallest element in the first position, it satisfies b(1)<b(2)<b(3)<... . A cycle is said to be even if it has an even number of entries. For example, the permutation (18)(2347)(569) has 2 increasing even cycles.
1, 1, 1, 1, 3, 3, 14, 7, 3, 70, 35, 15, 419, 226, 60, 15, 2933, 1582, 420, 105, 23421, 12741, 3423, 630, 105, 210789, 114669, 30807, 5670, 945, 2108144, 1144921, 311160, 55755, 7875, 945, 23189584, 12594131, 3422760, 613305, 86625, 10395, 278279165, 151125052, 41041968, 7429290, 1001385, 114345, 10395
0,5
E.g.f.: G(t,z)=exp((t-1)(cosh z - 1))/(1-z).
The 5-variate e.g.f. H(x,y,u,v,z) of permutations with respect to size (marked by z), number of increasing odd cycles (marked by x), number of increasing even cycles (marked by y), number of nonincreasing odd cycles (marked by u), and number of nonincreasing even cycles (marked by v), is given by
H(x,y,u,v,z)=exp(((x-u)sinh z + (y-v)(cosh z - 1))*(1+z)^{(u-v)/2}/(1-z)^{(u+v)/2}.
We have: G(t,z)=H(1,t,1,1,z).
T(3,1)=3 because we have (1)(23), (12)(3), and (13)(2).
T(4,2)=3 because we have (12)(34), (13)(24), and (14)(23).
Triangle starts:
1;
1;
1,1;
3,3;
14,7,3;
70,35,15;
g := exp((t-1)*(cosh(z)-1))/(1-z): gser := simplify(series(g, z = 0, 15)): for n from 0 to 12 do P[n] := sort(expand(factorial(n)*coeff(gser, z, n))) end do: for n from 0 to 12 do seq(coeff(P[n], t, k), k = 0 .. floor((1/2)*n)) end do; # yields sequence in triangular form
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Emeric Deutsch (deutsch(AT)duke.poly.edu), Feb 27 2011
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