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A186366
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Triangle read by rows: T(n,k) is the number of cycle-up-down permutations of {1,2,...,n} having k cycles (1<=k<=n).
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8
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1, 1, 1, 1, 3, 1, 2, 7, 6, 1, 5, 20, 25, 10, 1, 16, 70, 105, 65, 15, 1, 61, 287, 490, 385, 140, 21, 1, 272, 1356, 2548, 2345, 1120, 266, 28, 1, 1385, 7248, 14698, 15204, 8715, 2772, 462, 36, 1, 7936, 43280, 93420, 105880, 69405, 26985, 6090, 750, 45, 1, 50521, 285571, 649715, 793210, 577225, 260337, 72765, 12210, 1155, 55, 1
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OFFSET
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1,5
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COMMENTS
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A permutation is said to be cycle-up-down if it is a product of up-down cycles. A cycle (b(1), b(2), ...) is said to be up-down if, when written with its smallest element in the first position, it satisfies b(1) < b(2) > b(3) < ... .
Sum of entries in row n is A000111(n+1) (the Euler or up-down numbers).
The row generating polynomial of row n is c_n(t) from the Johnson reference (p. 127).
Sum_{k=1..n} k*T(n,k) = A186367(n).
With a different offset, appears to be the same as the table of sub-permutations of a class of Andre permutations given by Disanto. - Peter Bala, Feb 11 2012. That is, this triangle appears to be identical to the triangle giving the number of binary increasing trees with n nodes and a "min-path" of length k. - N. J. A. Sloane, May 12 2012
An apparent signed version is presented on p. 6 of the Csikvari preprint, related to graph polynomials of the complete graphs K_n. - Tom Copeland, Jan 17 2017
The trivariate e.g.f. below may be expressed as H = [e^(-z) e^(xz) / (1-sin (z))]^t = [e^(z*(p.(x)-1))]^t = [e^(z*(p.(x-1)))]^t, where (p.(x))^n = p_n(x) are a sequence of Appell polynomials. For t=m, an integer, the formalism of A248120 related to the Hirzebruch criterion for convolutions applies and that of the Scott and Sokal preprint (see eqn. 3.1 on p. 10 and eqn. 3.62 on p. 24). - Tom Copeland, Jan 17 2017
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LINKS
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FORMULA
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E.g.f.: 1/(1-sin(z))^t.
The trivariate e.g.f. H(t,s,z) of the cycle-up-down permutations of {1,2,...,n} with respect to size (marked by z), number of cycles (marked by t), and number of fixed points (marked by x) is given by H(t,x,z)=exp((x-1)tz)/(1-sin z)^t.
T(n,m) = 2*sum(j=floor((n+m)/2)..n, (stirling1(2*j-n,m)*(-1)^(m-n)*sum(i=0..(2*j-n)/2, (2*i-2*j+n)^n*binomial(2*j-n,i)*(-1)^(j-i)))/(2^(2*j-n)*(2*j-n)!)). - Vladimir Kruchinin, Mar 26 2013
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EXAMPLE
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T(3,2)=3 because we have (1)(23), (12)(3), and (13)(2).
T(4,3)=6 because we have (1)(2)(34), (1)(23)(4), (1)(24)(3), (12)(3)(4), (13)(2)(4), and (14)(2)(3).
Triangle starts:
1;
1, 1;
1, 3, 1;
2, 7, 6, 1;
5, 20, 25, 10, 1;
16, 70, 105, 65, 15, 1;
61, 287, 490, 385, 140, 21, 1;
272, 1356, 2548, 2345, 1120, 266, 28, 1;
1385, 7248, 14698, 15204, 8715, 2772, 462, 36, 1;
...
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MAPLE
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G := 1/(1-sin(z))^t: Gser := simplify(series(G, z = 0, 16)): for n to 11 do P[n] := sort(expand(factorial(n)*coeff(Gser, z, n))) end do: for n from 0 to 11 do seq(coeff(P[n], t, j), j = 1 .. n) end do; # yields sequence in triangular form
# second Maple program:
b:= proc(u, o) option remember; `if`(u+o=0, 1,
add(b(o-1+j, u-j), j=1..u))
end:
g:= proc(n) option remember; expand(`if`(n=0, 1,
add(g(n-j)*binomial(n-1, j-1)*x*b(j-1, 0), j=1..n)))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(g(n)):
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MATHEMATICA
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rows = 12;
a111[n_] := If[EvenQ[n], Abs[EulerE[n]], Abs[(2^(n+1)*(2^(n+1) - 1)* BernoulliB[n+1])/(n+1)]];
BellMatrix[f_, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len-1}, {k, 0, len-1}]];
B = BellMatrix[a111, rows];
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PROG
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(Maxima)
T(n, m):=2*sum((stirling1(2*j-n, m)*(-1)^(m-n)*sum((2*i-2*j+n)^n*binomial(2*j-n, i)*(-1)^(j-i), i, 0, (2*j-n)/2))/(2^(2*j-n)*(2*j-n)!), j, floor((n+m)/2), n); /* Vladimir Kruchinin, Mar 26 2013 */
(Sage) # uses[bell_matrix from A264428]
# Adds a column 1, 0, 0, 0, ... at the left side of the triangle.
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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