A102233 Number of preferential arrangements of n labeled elements when at least k=3 elements per rank are required.

1, 0, 0, 1, 1, 1, 21, 71, 183, 2101, 13513, 64285, 629949, 5762615, 41992107, 427215283, 4789958371, 47283346849, 540921904725, 6980052633257, 85901272312905, 1129338979629643, 16398293425501375, 238339738265039119, 3588600147767147775, 58124879519314730741

Offset: 0

Author

Thomas Wieder, Jan 01 2005

Keywords

• nonn

Comments

The labeled case for at least k=2 elements per rank is given by A032032 = Partition n labeled elements into sets of sizes of at least 2 and order the sets. The unlabeled case for at least k=3 elements per rank is given by A000930 = A Lamé sequence of higher order. The unlabeled case for at least k=2 elements per rank is given by A000045 = Fibonacci numbers.

With m = floor(n/3), a(n) is the number of ways to distribute n different toys to m numbered children such that each child receiving a toy gets at least three toys and, if child k gets no toys, then each child numbered higher than k also gets no toys. Furthermore, a(n)= row sums of triangle A200092 for n>=3. - Dennis P. Walsh, Apr 15 2013

Row sums of triangle A200092. - Dennis P. Walsh, Apr 15 2013

Examples

			Let 1,2,3,4,5,6 denote six labeled elements. Let | denote a separation between two ranks. E.g., if elements 1,2 and 3 are on rank (also called level) one and elements 3,4 and 5 are on rank two, then we have the ranking 123|456.
For n=9 we have a(9)=2101 rankings. The order within a rank does not count. Six examples are: 123|456|789; 123456789; 12345|6789; 129|345678; 1235|46789; 789|123456.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..400
• Vladimir Kruchinin, D. V. Kruchinin, Composita and their properties, arXiv:1103.2582 [math.CO], 2011-2013.
• I. Mezo, Periodicity of the last digits of some combinatorial sequences, arXiv preprint arXiv:1308.1637 [math.CO], 2013 and J. Int. Seq. 17 (2014) #14.1.1.

Crossrefs

Cf. A000640, A102232, A032032, A232475.

Cf. column k=3 of A245732.

Programs

Maple

seq (n! *coeff (series (1- (z^2-2*exp(z)+2+2*z) /(4-2*exp(z)+2*z+z^2), z=0, n+1), z, n), n=0..30);
with(combstruct): SeqSetL := [S, {S=Sequence(U), U=Set(Z, card >= 3)}, labeled]: seq(count(SeqSetL, size=j), j=0..23); # Zerinvary Lajos, Oct 19 2006
# third Maple program:
b:= proc(n) b(n):= `if`(n=0, 1, add(b(n-j)/j!, j=3..n)) end:
a:= n-> n!*b(n):
seq(a(n), n=0..30);  # Alois P. Heinz, Jul 29 2014

Mathematica

CoefficientList[Series[1-(x^2-2*E^x+2+2*x)/(4-2*E^x+2*x+x^2), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Sep 29 2013 *)
b[n_] := b[n] = If[n==0, 1, Sum[b[n-j]/j!, {j, 3, n}]]; a[n_] := n!*b[n]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jan 31 2016, after Alois P. Heinz *)

PARI

z='z+O('z^66); Vec(serlaplace( 1-(z^2-2*exp(z)+2+2*z) / (4-2*exp(z)+2*z+z^2) ) ) \\ Joerg Arndt, Apr 16 2013

Formula

E.g.f.: 1-(z^2-2*exp(z)+2+2*z)/(4-2*exp(z)+2*z+z^2).

a(n) = n! * sum(m=1..n, sum(k=0..m, k!*(-1)^(m-k) *binomial(m,k) *sum(i=0..n-m, stirling2(i+k,k) *binomial(m-k,n-m-i) *2^(-n+m+i) /(i+k)!))); a(0)=1. - Vladimir Kruchinin, Feb 01 2011

a(n) ~ 2*n!/((2+r^2)*r^(n+1)), where r = 1.56811999239... is the root of the equation 4+2*r+r^2 = 2*exp(r). - Vaclav Kotesovec, Sep 29 2013

a(0) = 1; a(n) = Sum_{k=3..n} binomial(n,k) * a(n-k). - Ilya Gutkovskiy, Feb 09 2020

E.g.f.: 1/(2 + x + x^2/2 - exp(x)). - Christian Sievers, Oct 27 2024

Extensions

a(0) changed to 1 at the suggestion of Zerinvary Lajos, Oct 26 2006