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
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.
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.
