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
References
- 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
- 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.
Programs
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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
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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 *) -
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 */
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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
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
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
Extensions
More terms from Christian G. Bower, Oct 15 1999
Comments