A006154 Number of labeled ordered partitions of an n-set into odd parts. (Formerly M1792)

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

Offset

0,3

Comments

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

References

Getu, S.; Shapiro, L. W.; Combinatorial view of the composition of functions. Ars Combin. 10 (1980), 131-145.

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

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Philippe Flajolet, Symbolic Enumerative Combinatorics and Complex Asymptotic Analysis, Algorithms Seminar 2000-2001, F. Chyzak (ed.), INRIA, (2002), pp. 161-170.

Formula

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

Maple

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

Mathematica

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 *)

Prog

(PARI) 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
(Maxima) 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 */

Crossrefs

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

Sequence in context: A097900 A198891 A000153 * A352435 A000987 A006957

Adjacent sequences: A006151 A006152 A006153 * A006155 A006156 A006157

Keyword

nonn,easy,nice

Author

Simon Plouffe

Extensions

More terms from Christian G. Bower, Oct 15 1999

Status

approved