A113485 Number of partitions of [n] avoiding the pattern 12/34.

1, 2, 5, 14, 41, 122, 367, 1114, 3423, 10670, 33841, 109398, 361045, 1217346, 4195267, 14775986, 53172411, 195396310, 732806677, 2802898190, 10926431393, 43381582538, 175311002903, 720640632074, 3011495745175, 12786738800254

Offset: 1

Author

Adam Goyt, Jan 09 2006

Keywords

• nonn

Comments

The first sum in the formula counts those partitions with a single block of size at least 3. The second sum counts those partitions with blocks of size at most 2. It's easy to see that to avoid 12/34 a partition cannot contain more than one block of size at least 3.

The elements shown satisfy the hypergeometric recurrence 2*a(n) -10*a(n-1) +(-n+13)*a(n-2) +2*(2*n+1)*a(n-3) +3*(-n-5)*a(n-4) +4*(-n+6)*a(n-5) +4*(n-5)*a(n-6)=0. - R. J. Mathar, Jan 25 2013

Examples

			For n=1,2,3 a(n)=B_n, where B_n is the n-th Bell number, since there aren't enough distinct elements for such a partition to contain a copy of 12/34. By a similar argument a(4)=B_4-1=14.
		

References

• M. Klazar, Counting Pattern-free Set Partitions I: A Generalization of Stirling Numbers of the Second Kind, Europ. J. Combinatorics, Vol. 21 (2000), pp. 367-378.

Links

• G. C. Greubel, Table of n, a(n) for n = 1..880

Crossrefs

Cf. A084261.

Programs

Mathematica

Table[Sum[Sum[(k+1)^2 Binomial[n, 2k+p] k!, {k, 0, Floor[(n-p)/2]}], {p, 3, n}]+Sum[Binomial[n, 2k] k!, {k, 0, Floor[n/2]}], {n, 1, 31}]

PARI

a(n)=sum(p=3,n, sum(k=0,(n-p)\2, binomial(n,2*k+p)*(k+1)^2*k!)) + sum(k=0,n\2, binomial(n,2*k)) \\ Charles R Greathouse IV, Mar 12 2017

Formula

a(n) = Sum[Sum[(k+1)^2 binomial[n, 2k+p] k!, {k, 0, Floor[(n-p)/2]}], {p, 3, n}] + Sum[binomial[n, 2k] k!, {k, 0, Floor[n/2]}].

From Vaclav Kotesovec, Jun 10 2019: (Start)

Recurrence: 2*(n^2-8*n+13)*a(n) = 2*(4*n^2-31*n+43)*a(n-1) + (n^3-17*n^2+87*n-91)*a(n-2) - (3*n^3-27*n^2+78*n-64)*a(n-3) + 2*(n-3)*(n^2-6*n+6)*a(n-4).

a(n) ~ sqrt(Pi) * exp(sqrt(2*n) - n/2 - 1/2) * n^(n/2 + 1) / 2^(n/2 + 3/2) * (1 + 4*sqrt(2)/(3*sqrt(n))). (End)