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A102233 Number of preferential arrangements of n labeled elements when at least k=3 elements per rank are required. 11

%I #62 Mar 28 2020 15:03:16

%S 1,0,0,1,1,1,21,71,183,2101,13513,64285,629949,5762615,41992107,

%T 427215283,4789958371,47283346849,540921904725,6980052633257,

%U 85901272312905,1129338979629643,16398293425501375,238339738265039119,3588600147767147775,58124879519314730741

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

%C 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.

%C 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

%C Row sums of triangle A200092. - _Dennis P. Walsh_, Apr 15 2013

%H Alois P. Heinz, <a href="/A102233/b102233.txt">Table of n, a(n) for n = 0..400</a>

%H Vladimir Kruchinin, D. V. Kruchinin, <a href="http://arxiv.org/abs/1103.2582">Composita and their properties</a>, arXiv:1103.2582 [math.CO], 2011-2013.

%H I. Mezo, <a href="http://arxiv.org/abs/1308.1637">Periodicity of the last digits of some combinatorial sequences</a>, arXiv preprint arXiv:1308.1637 [math.CO], 2013 and <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Mezo/mezo19.html">J. Int. Seq. 17 (2014) #14.1.1</a>.

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

%F 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

%F 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

%F a(0) = 1; a(n) = Sum_{k=3..n} binomial(n,k) * a(n-k). - _Ilya Gutkovskiy_, Feb 09 2020

%e 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.

%e 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.

%p 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);

%p 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

%p # third Maple program:

%p b:= proc(n) b(n):= `if`(n=0, 1, add(b(n-j)/j!, j=3..n)) end:

%p a:= n-> n!*b(n):

%p seq(a(n), n=0..30); # _Alois P. Heinz_, Jul 29 2014

%t 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 *)

%t 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_ *)

%o (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

%Y Cf. A000640, A102232, A032032, A232475.

%Y Cf. column k=3 of A245732.

%K nonn

%O 0,7

%A _Thomas Wieder_, Jan 01 2005

%E a(0) changed to 1 at the suggestion of _Zerinvary Lajos_, Oct 26 2006

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Last modified April 24 20:08 EDT 2024. Contains 371963 sequences. (Running on oeis4.)