A190826 Number of permutations of 3 copies of 1..n introduced in order 1..n with no element equal to another within a distance of 1.
1, 0, 1, 29, 1721, 163386, 22831355, 4420321081, 1133879136649, 372419001449076, 152466248712342181, 76134462292157828285, 45552714996556390334921, 32173493282909179882613934, 26487410329744429030530295991, 25143126122564855343240882599761, 27260957330891104469298062949026065
Offset: 0
Keywords
Examples
Some of the a(3) = 29 solutions for n=3: 123232131, 123121323, 123123213, 123212313, 123213123, 121323132, 123132312, 123123123, 123231213, 121323123, 121321323, 121312323, 121323231, 123231321, 121313232, 123132321, ...
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..223 (terms 0..101 from Andrew Woods)
- H. Eriksson and A. Martin, Enumeration of Carlitz multipermutations, arXiv:1702.04177 [math.CO], 2017.
- R. J. Mathar, A class of multinomial permutations avoiding object clusters, vixra:1511.0015 (2015), sequence M_{c,3}/3!.
Programs
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Magma
B:=Binomial; f:= func< n,j | (&+[B(n,k)*B(2*k,j)*(-3)^(k-j): k in [Ceiling(j/2)..n]]) >; A190826:= func< n | (-1/2)^n*(&+[Factorial(j)*B(n+j,j)*f(n,j): j in [0..2*n]]) >; [A190826(n): n in [0..30]]; // G. C. Greubel, Sep 22 2023
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Mathematica
a[n_]:= 1/(6^n*n!)*Sum[(n+j)! Sum[Binomial[n,k] Binomial[2k,j] (-3)^(n+k-j), {k, Ceiling[j/2], n}], {j,0,2n}]; Array[a, 16, 0] (* Jean-François Alcover, Jul 22 2017, after Tani Akinari's code for A193638 *) -
SageMath
b=binomial; def f(j,n): return sum(b(n,k)*b(2*k,j)*(-3)^(k-j) for k in range((j//2),n+1)) def A190826(n): return (-1/2)^n*sum(factorial(j)*b(n+j,j)*f(j,n) for j in range(2*n+1)) [A190826(n) for n in range(31)] # G. C. Greubel, Sep 22 2023
Formula
a(n) = A193624(n)/(6^n * n!), for n >= 1.
a(n) = A193638(n)/n!, for n >= 1.
a(n) = A192990(binomial(n+2,3)) / (6^n * n!), for n >= 1.
2*a(n) -3*(3*n^2-3*n+4)*a(n-1) +2*(9*n^2-42*n+47)*a(n-2) +8*(3*n-7)*a(n-3) -8*a(n-4) = 0. - R. J. Mathar, May 23 2014
a(n) = (1/(6^n * n!)) * Sum_{j=0..2*n} Sum_{k=ceiling(j/2)..n} (n+j)! * binomial(2*k, j) * binomial(n, k) * (-3)^(n+k-j). - Jean-François Alcover, Jul 22 2017
a(n) ~ 3^(2*n + 1/2) * n^(2*n) / (2^n * exp(2*n + 2)). - Vaclav Kotesovec, Nov 24 2018
Extensions
a(0)=1 prepended by Alois P. Heinz, Jul 22 2017
Comments