%I
%S 0,1,0,3,14,72,468,3453,28782,267831,2752828,30984336,379125192,
%T 5011756625,71190365580,1081514329155,17499480412746,300473929597320,
%U 5457031426340748,104520033700333069,2105651342251571562
%N Number of successions in all the permutations p of [n] such that p(1)=1 and having no 3sequences. A succession of a permutation p is a position i such that p(i +1)  p(i) = 1.
%C From _Emeric Deutsch_, Sep 07 2010: (Start)
%C a(n) is also the number of fixed points in all those permutations of [n1] that have no adjacent fixed points. Example: a(4)=3 because in 132, 213, 231, 312, 321 we have 1+1+0+0+1 fixed points.
%C a(n) is also the number of permutations of [n] having exactly 1 pair of adjacent fixed points. Example: a(4)=3 because we have 1243, 4231, and 2134.
%C (End)
%D Wayne M. Dymacek, Isaac Lambert and Kyle Parsons, Arithmetic Progressions in Permutations, http://math.ku.edu/~ilambert/CN.pdf, 2012.  From _N. J. A. Sloane_, Sep 15 2012
%F a(n) = Sum_{k>=0} k*A180186(n,k).
%F a(n) = Sum_{k=0..floor(n/2)} k*binomial(nk,k)*d(n1k), where d(j)=A000166(j) are the derangement numbers.
%e a(4)=3 because in 1*243, 1324, 13*42, 142*3, 1432 we have 3 successions (marked *).
%p d[0] := 1: for n to 51 do d[n] := n*d[n1]+(1)^n end do: seq(sum(k*binomial(nk, k)*d[n1k], k = 0 .. floor((1/2)*n)), n = 1 .. 22);
%t a[0] = 1; a[n_] := a[n] = n*a[n  1] + (1)^n; f[n_] := Sum[k*Binomial[n  k, k]*a[n  k  1], {k, 0, n/2}]; Array[f, 21] (* _Robert G. Wilson v_, Apr 01 2011 *)
%t a[n_] := Sum[k Binomial[n  k, k] Subfactorial[n  k  1], {k, 0, n/2}];
%t a /@ Range[1, 21] (* _JeanFrançois Alcover_, Oct 29 2019 *)
%Y Cf. A000166, A180186.
%Y A column of A216718.  _N. J. A. Sloane_, Sep 15 2012
%K nonn
%O 1,4
%A _Emeric Deutsch_, Sep 06 2010
