%I
%S 1,2,4,10,28,86,282,984,3630,14138,57904,248854,1118554,5246980,
%T 25619018,129961850,683561488,3722029314,20946195078,121671375312,
%U 728511702462,4491224518274,28475638336144,185499720543262,1240358846060122,8505894459387628,59771243719783410
%N The number of ways to write an nbit binary string and then define each run of ones as an element in an equivalence relation, and each run of zeros as an element in a second equivalence relation.
%C Included are the cases in which there are no zeros or no ones, producing an empty relation.
%H Alois P. Heinz, <a href="/A253409/b253409.txt">Table of n, a(n) for n = 0..647</a>
%F a(n) = 2 * Sum_{k=1..ceiling(n/2)} C(n1,2k1)*Bell(k)^2 + C(n1,2k2)*Bell(k)*Bell(k1), where C(x,y) refers to binomial coefficients and Bell(x) refers to Bell numbers (A000110).
%e For n = 3, taking 3bit binary strings and replacing zeros with ABC... and ones with 123... to represent equivalence relations, we have a(3) = 10 labeledrun binary strings: AAA, AA1, A1A, A1B, 1AA, A11, 11A, 111, 1A1, 1A2.
%t Table[2 * Sum[Binomial[n1,2k1] * BellB[k]^2 + Binomial[n1,2k2] * BellB[k] * BellB[k1],{k,1,Ceiling[n/2]}],{n,1,30}] (* _Vaclav Kotesovec_, Jan 08 2015 after _Andrew Woods_ *)
%Y Cf. A000110, A247100.
%K nonn
%O 0,2
%A _Andrew Woods_, Jan 01 2015
%E a(0)=1 prepended by _Alois P. Heinz_, Aug 08 2015
