A186756 - OEIS

%I #17 Nov 28 2016 09:06:04

%S 1,1,2,5,1,15,9,52,68,203,507,10,877,3918,245,4140,32057,4123,21147,

%T 280700,60753,280,115975,2645611,853914,13300,678570,26917867,

%U 11923428,396935,4213597,295934526,169127222,9710855,15400,27644437,3513447546,2469452843,215274774,1201200

%N Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,n} having k nonincreasing cycles (0<=k<=floor(n/3)). 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) < ... .

%C Row n contains 1 + floor(n/3) entries.

%C Sum of entries in row n is n!.

%C T(n,0) = A000110(n) (the Bell numbers).

%C Sum_{k=0..n} k*T(n,k) = A121633(n).

%H Alois P. Heinz, <a href="/A186756/b186756.txt">Rows n = 0..250, flattened</a>

%F E.g.f.: G(t,z) = exp((1-t)(exp(z)-1))/(1-z)^t.

%F The 4-variate e.g.f. H(u,v,w,z) of the permutations of {1,2,...,n} with respect to size (marked by z), number of fixed points (marked by u), number of increasing cycles of length >=2 (marked by v), and number of nonincreasing cycles (marked by w) is given by H(u,v,w,z)=exp(uz+v(exp(z)-1-z)+w(1-exp(z))/(1-z)^w. Remark: the nonincreasing cycles are necessarily of length >=3. We have: G(t,z) = H(1,1,t,z).

%e T(3,0)=5 because we have (1)(2)(3), (1)(23), (12)(3), (13)(2), and (123).

%e T(3,1)=1 because we have (132).

%e T(4,1)=9 because we have (1)(243), (1432), (142)(3), (132)(4), (1342), (1423), (1243), (143)(2), and (1324).

%e Triangle starts:

%e     1;

%e     1;

%e     2;

%e     5,   1;

%e    15,   9;

%e    52,  68;

%e   203, 507, 10;

%p G := exp((1-t)*(exp(z)-1))/(1-z)^t: Gser := simplify(series(G, z = 0, 16)): for n from 0 to 13 do P[n] := sort(expand(factorial(n)*coeff(Gser, z, n))) end do: for n from 0 to 13 do seq(coeff(P[n], t, j), j = 0 .. floor((1/3)*n)) end do; # yields sequence in triangular form

%p # second Maple program:

%p b:= proc(n) option remember; expand(`if`(n=0, 1, add(

%p       (1+x*(i!-1))*b(n-i-1)*binomial(n-1, i), i=0..n-1)))

%p     end:

%p T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n)):

%p seq(T(n), n=0..15);  # _Alois P. Heinz_, Sep 26 2016

%t b[n_] := b[n] = Expand[If[n==0, 1, Sum[(1+x*(i!-1))*b[n-i-1]*Binomial[n-1, i], {i, 0, n-1}]]]; T[n_] := Function [p, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][b[n]]; Table[T[n], {n, 0, 15}] // Flatten (* _Jean-François Alcover_, Nov 28 2016 after _Alois P. Heinz_ *)

%Y Cf. A000110, A121633, A186754, A186755, A186757, A186758, A186759, A186760.

%K nonn,tabf

%O 0,3

%A _Emeric Deutsch_, Feb 26 2011