A250029 Maximum number of binary strings with symmetrically partitioned n 1's and n 0's, counted up to isomorphism.

1, 1, 1, 4, 9, 16, 36, 144, 400, 900, 3600, 11025, 28224, 78400, 254016, 705600, 2286144, 6350400, 25401600, 85377600, 250905600, 768398400, 3073593600, 10600761600, 32464832400, 129859329600, 456536705625

Offset: 0

Author

Andrei Cretu, Nov 11 2014

Keywords

• nonn

Comments

The number of binary strings, counted up to isomorphism, that can be constructed by taking an equal number (n) of 0's and 1's and partitioning both the 0's and the 1's into m runs using the same partition, can be written as:

dualseq[partition[n]]=m!^2/(Prod_j(m_j!))^2

where m_j is the multiplicity of runs of length j.

The numbers satisfy the relations Sum_j(m_j)=m, Sum_j(j*m_j)=n.

The strings obtained in this manner are a subset of those in A247651.

Both the finest and coarsest partitions of n minimize dualseq[partition[n]]. In this sense, dualseq[partition[n]] is another relative measure of the complexity of the partition and the associated binary strings.

a[n] is the number of strings, counted up to isomorphism, that can be generated based on the partition that maximizes dualseq[partition[n]].

Examples

			n=0 gives the empty string.
n=1 and the only possible partition generate 01 (and the isomorphic 10).
For n=2, both possible partitions generate, up to isomorphism, 1 string, 0011 (1100), and respectively 0101 (1010).
For n=3, the optimal partition is {1,2}, generating, up to isomorphism, 4 strings: 001011 (110100), 001101 (110010), 010011 (101100) and 011001 (100110).
For n=4, the optimal partition is {1,1,2}, generating, up to isomorphism, 9 strings: 00101011 (11010100), 00101101 (11010010), 00110101 (11001010), 01001011 (10110100), 01001101 (10110010), 01010011 (10101100), 01011001 (10100110), 01100101 (10011010) and 01101001 (10010110).
		

Crossrefs

Cf. A247651, A046952, A092266, A136404, A176471, A065886.

Programs

Mathematica

dualseq[p_]:=Factorial[Length[p]]^2/Apply[Times,Map[Factorial[Count[p,#1]]&,Range[Max[Length[p]]]]]^2
a[n_]:=Max[Map[dualseq,IntegerPartitions[n]]]
Table[a[n],{n,0,25}] (* after A130670 *)

Formula

a[n]=Max[m!^2/(Prod_j(m_j!))^2] where Sum_j(m_j)=m, Sum_j(j*m_j)=n, over all partitions of n.