login
Number of set partitions of [n] such that at least one pair of consecutive blocks (b,b+1) exists having no pair of consecutive numbers (i,i+1) with i member of b and i+1 member of b+1.
4

%I #12 Feb 02 2017 09:15:15

%S 0,0,0,0,1,9,58,341,1983,11776,72345,462173,3075894,21330762,

%T 154050330,1157493707,9037925277,73244123107,615295131046,

%U 5351329029624,48126530239366,447043890866154,4284293705043796,42317095568379559,430355360965092107,4501973706497500364

%N Number of set partitions of [n] such that at least one pair of consecutive blocks (b,b+1) exists having no pair of consecutive numbers (i,i+1) with i member of b and i+1 member of b+1.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Partition_of_a_set">Partition of a set</a>

%F a(n) = A000110(n) - A271270(n).

%e a(4) = 1: 13|2|4.

%e a(5) = 9: 124|3|5, 134|2|5, 135|2|4, 13|25|4, 13|2|45, 13|2|4|5, 14|23|5, 14|2|3|5, 1|24|3|5.

%p b:= proc(n, i, m, l) option remember; `if`(n=0,

%p `if`(l=[] or {l[]}={1}, 1, 0), add(b(n-1, j, max(m, j),

%p `if`(j=m+1, `if`(j=i+1, [l[],1], [l[],0]),

%p `if`(j=i+1, subsop(j=1, l), l))), j=1..m+1))

%p end:

%p a:= n-> combinat[bell](n)-b(n, 0$2, []):

%p seq(a(n), n=0..18);

%t b[n_, i_, m_, l_] := b[n, i, m, l] = If[n == 0, If[Union[l, {1}] == {1}, 1, 0], Sum[b[n-1, j, Max[m, j], If[j == m+1, Join[l, If[j == i+1, {1}, {0}] ], If[j == i+1, ReplacePart[l, j -> 1], l]]], {j, 1, m+1}]]; a[n_] := BellB[n] - b[n, 0, 0, {}]; Table[a[n], {n, 0, 18}] (* _Jean-François Alcover_, Feb 02 2017, translated from Maple *)

%Y Cf. A000110, A185982, A271270, A271273, A272065.

%K nonn

%O 0,6

%A _Alois P. Heinz_, Apr 03 2016