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A162971
Triangle read by rows: T(n,k) is number of non-derangement permutations of {1,2,...,n} having k cycles (1 <= k <= n).
2
1, 0, 1, 0, 3, 1, 0, 8, 6, 1, 0, 30, 35, 10, 1, 0, 144, 210, 85, 15, 1, 0, 840, 1414, 735, 175, 21, 1, 0, 5760, 10752, 6664, 1960, 322, 28, 1, 0, 45360, 91692, 64764, 22449, 4536, 546, 36, 1, 0, 403200, 869040, 679580, 268380, 63273, 9450, 870, 45, 1, 0, 3991680, 9074736, 7704180, 3382280, 902055, 157773, 18150, 1320, 55, 1
OFFSET
1,5
COMMENTS
Sum of entries in row n = A002467(n) (the number of non-derangement permutations of {1,2,...,n}).
T(n,2) = n*(n-2)! = A001048(n-1) for n>=3.
Sum_{k=1..n} k*T(n,k) = A162972(n).
LINKS
FORMULA
E.g.f.: G(t,z) = (1-exp(-tz))/(1-z)^t.
EXAMPLE
T(4,2) = 8 because we have (1)(234), (1)(243), (134)(2), (143)(2), (124)(3), (142)(3), (123)(4), and (132)(4).
Triangle starts:
1;
0, 1;
0, 3, 1;
0, 8, 6, 1;
0, 30, 35, 10, 1;
0, 144, 210, 85, 15, 1;
...
MAPLE
G := (1-exp(-t*z))/(1-z)^t: Gser := simplify(series(G, z = 0, 15)): for n to 11 do P[n] := sort(expand(factorial(n)*coeff(Gser, z, n))) end do: for n to 11 do seq(coeff(P[n], t, j), j = 1 .. n) end do; # yields sequence in triangular form
# second Maple program:
b:= proc(n, t) option remember; `if`(n=0, t, add(expand((j-1)!*
b(n-j, `if`(j=1, 1, t))*x)*binomial(n-1, j-1), j=1..n))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n, 0)):
seq(T(n), n=1..12); # Alois P. Heinz, Aug 15 2023
MATHEMATICA
b[n_, t_] := b[n, t] = If[n == 0, t, Sum[Expand[(j - 1)!*b[n - j, If[j == 1, 1, t]]*x]*Binomial[n - 1, j - 1], {j, 1, n}]];
T[n_] := CoefficientList[b[n, 0]/x, x];
Table[T[n], {n, 1, 12}] // Flatten (* Jean-François Alcover, Apr 04 2024, after Alois P. Heinz *)
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Emeric Deutsch, Jul 22 2009
STATUS
approved