A193474 -id:A193474 - cp's OEIS frontend

A006154 Number of labeled ordered partitions of an n-set into odd parts.

1, 1, 2, 7, 32, 181, 1232, 9787, 88832, 907081, 10291712, 128445967, 1748805632, 25794366781, 409725396992, 6973071372547, 126585529106432, 2441591202059281, 49863806091395072, 1074927056650469527, 24392086908129247232, 581176736647853024581

Offset: 0

Simon Plouffe

Conjecture: taking the sequence modulo an integer k gives an eventually periodic sequence. For example, the sequence taken modulo 10 is [1, 1, 2, 7, 2, 1, 2, 7, 2, 1, 2, 7, 2, ...], with an apparent period [1, 2, 7, 2] beginning at a(1), of length 4. Cf. A000670. - Peter Bala, Apr 12 2023

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Farid Aliniaeifard and Shu Xiao Li, Peak algebra in noncommuting variables, arXiv:2506.12868 [math.CO], 2025. See pp. 2, 9, 45.
Philippe Flajolet, Symbolic Enumerative Combinatorics and Complex Asymptotic Analysis, Algorithms Seminar 2000-2001, F. Chyzak (ed.), INRIA, (2002), pp. 161-170.
S. Getu and L. W. Shapiro, Combinatorial view of the composition of functions, Ars Combin. 10 (1980), 131-145. (Annotated scanned copy)
Prabha Sivaraman Nair and Rejikumar Karunakaran, On k-Fibonacci Brousseau Sums, J. Int. Seq. (2024) Art. No. 24.6.4. See p. 8.

Cf. A000045, A000111, A000670, A003724, A193474, A196776.

readlib(coeftayl):
with(combinat, bell);
A:=series(1/(1-sinh(x)),x,20);
G(x):=A : f[0]:=G(x): for n from 0 to 21 do f[n]:=coeftayl(G(x), x=0, n);;
p[n]:=f[n]*((n)!) od: x:=0:seq(p[n], n=0..20); # Sergei N. Gladkovskii, Jun 01 2012
# second Maple program:
a:= proc(n) option remember; `if`(n=0, 1, add((i->
      a(n-i)*binomial(n, i))(2*j+1), j=0..(n-1)/2))
    end:
seq(a(n), n=0..23);  # Alois P. Heinz, Feb 01 2022

a[n_] := Sum[ (-1)^i*(k - 2*i)^n*Binomial[k, i]/2^k, {k, 1, n}, {i, 0, k}]; a[0] = 1; Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Dec 07 2011, after Vladimir Kruchinin *)
With[{nn=20},CoefficientList[Series[1/(1-Sinh[x]),{x,0,nn}],x]Range[0,nn]!] (* Harvey P. Dale, Nov 16 2012 *)

a(n):=sum(sum((-1)^i*(k-2*i)^n*binomial(k,i),i,0,k)/2^k,k,1,n); /* Vladimir Kruchinin, May 28 2011 */

a(n)=if(n<2,n>=0,sum(k=1,ceil(n/2),binomial(n,2*k-1)*a(n-2*k+1))) \\ Ralf Stephan

E.g.f.: 1/(1 - sinh(x)).
With alternating signs, e.g.f.: 1/(1+sinh(x)). - Ralf Stephan, Apr 29 2004
a(0) = a(1) = 1, a(n) = Sum_{k=1..ceiling(n/2)} C(n,2*k-1)*a(n-2*k+1). - Ralf Stephan, Apr 29 2004
a(n) ~ (sqrt(2)/2)*n!/log(1+sqrt(2))^(n+1). - Conjectured by Simon Plouffe, Feb 17 2007.
From Andrew Hone, Feb 22 2007: (Start)
This formula can be proved using the techniques in the article by Philippe Flajolet (see links) [see Theorem 5 and Table 2, noting that 1/(1-sinh(x)) just has a simple pole at x=log(1+sqrt(2))]. (End)
a(n) = Sum_{k=1..n} Sum_{i=0..k} (-1)^i*(k-2*i)^n*binomial(k,i)/2^k, n > 0, a(0)=1. - Vladimir Kruchinin, May 28 2011
Row sums (apart from a(0)) of A196776. - Peter Bala, Oct 06 2011
Row sums of A193474. - Peter Luschny, Oct 07 2011
a(n) = D^n(1/(1-x)) evaluated at x = 0, where D is the operator sqrt(1+x^2)*d/dx. Cf. A003724 and A000111. - Peter Bala, Dec 06 2011
From Sergei N. Gladkovskii, Jun 01 2012: (Start)
Let E(x) be the e.g.f., then
E(x) = -1/x + 1/(x*(1-x))+ x^3/((1-x)*((1-x)*G(0) - x^2)); G(k) = (2*k+2)*(2*k+3)+x^2-(2*k+2)*(2*k+3)*x^2/G(k+1); (continued fraction).
E(x) = -1/x + 1/(x*(1-x))+ x^3/((1-x)*((1-x)*G(0) - x^2)); G(k) = 8*k+6+x^2/(1 + (2*k+2)*(2*k+3)/G(k+1)); (continued fraction).
E(x) = 1/(1 - x*G(0)); G(k) = 1 + x^2/(2*(2*k+1)*(4*k+3) + 2*x^2*(2*k+1)*(4*k+3)/(-x^2 - 4*(k+1)*(4*k+5)/G(k+1))); (continued fraction).
(End).
E.g.f. 1/(1 - x*G(0)) where G(k) = 1 - x^2/( (2*k+1)*(2*k+3) - (2*k+1)*(2*k+3)^2/(2*k+3 - (2*k+2)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Oct 01 2012
O.g.f A(x) satisfies A(x) = 1 + ( A(x/(1-x))/(1-x) - A(x/(1+x))/(1+x) )/2. - Paul D. Hanna, Aug 19 2024

More terms from Christian G. Bower, Oct 15 1999

A196776 Triangle T(n,k) gives the number of ordered partitions of an n set into k odd-sized blocks.

Original entry on oeis.org

1, 0, 2, 1, 0, 6, 0, 8, 0, 24, 1, 0, 60, 0, 120, 0, 32, 0, 480, 0, 720, 1, 0, 546, 0, 4200, 0, 5040, 0, 128, 0, 8064, 0, 40320, 0, 40320, 1, 0, 4920, 0, 115920, 0, 423360, 0, 362880, 0, 512, 0, 130560, 0, 1693440, 0, 4838400, 0, 3628800

Offset: 1

Peter Bala, Oct 06 2011

See A136630 for the case of unordered partitions into odd-sized blocks. See A193474 for this triangle in row reverse form (but with an offset of 0).

			Triangle begins
.n\k.|..1....2....3....4.....5....6.....7
= = = = = = = = = = = = = = = = = = = = =
..1..|..1
..2..|..0....2
..3..|..1....0....6
..4..|..0....8....0...24
..5..|..1....0...60....0...120
..6..|..0...32....0..480.....0..720
..7..|..1....0..546....0..4200....0..5040
...
T(4,2) = 8: The 8 ordered partitions of the set {1,2,3,4} into 2 odd-sized blocks are {1}{2,3,4}, {2,3,4}{1}, {2}{1,3,4}, {1,3,4}{2}, {3}{1,2,4}, {1,2,4}{3}, {4}{1,2,3} and {1,2,3}{4}.
Example of recurrence relation: T(7,3) = 3^2*T(5,3) + 3*(3-1)*T(5,1) = 9*60 + 6*1 = 546.

Cf. A006154 (row sums), A136630, A162590, A193474 (row reverse).

T(n,k) = 1/(2^k)*sum {j = 0..k}(-1)^(k-j)*binomial(k,j)*(2*j-k)^n.
Recurrence: T(n+2,k) = k^2*T(n,k) + k*(k-1)*T(n,k-2).
E.g.f.: x*sinh(t)/(1-x*sinh(t)) = x*t + 2*x^2*t^2/2! + (x+6*x^3)*t^3/3! + (8*x^2+24*x^4)*t^4/4! + (x+60*x^3+120*x^5)*t^5/5! + ....
O.g.f. for column 2*k: (2*k)!*x^(2*k)/Product {j = 0..k} (1 - (2*j)^2*x^2).
O.g.f. for column 2*k+1: (2*k+1)!*x^(2*k+1)/Product {j = 0..k} (1 - (2*j+1)^2*x^2).
Let P denote Pascal's triangle A070318 and put M = 1/2*(P-P^-1). M is A162590 (see also A131047). Then the first column of (I-t*M)^-1 (apart from the initial 1) lists the row polynomials for the present triangle.
n-th row sum = A006154(n).
Row generating polynomials equal D^n(1/(1-x*t)) evaluated at x = 0, where D is the operator sqrt(1+x^2)*d/dx. Cf. A136630. - Peter Bala, Dec 06 2011

cp's OEIS Frontend

A006154 Number of labeled ordered partitions of an n-set into odd parts.

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