%I #14 Jul 10 2020 22:07:34
%S 1,4,20,100,525,2912,17052,105240,683100,4652340,33168850,246999480,
%T 1917186635,15480884720,129811538960,1128494172720,10155257740443,
%U 94465951576560,907162152191470,8982422995787780,91603484234843812
%N Number of partitions of {1...n} containing 3 pairs of consecutive integers, where each pair is counted within a block and a string of more than 2 consecutive integers are counted two at a time.
%H A. O. Munagi, <a href="http://dx.doi.org/10.1155/IJMMS.2005.451">Set partitions with successions and separations</a>, Int. J. Math. Math. Sci. (IJMMS) vol 2005 no 3 (2005) pp 451-463.
%F a(n) = binomial(n-1, 3)*Bell(n-4), the case r = 3 in the general case of r pairs: c(n, r) = binomial(n-1, r)*B(n-r-1).
%F O.g.f. for c(n,r) is exp(-1)*Sum(x^(r+1)/(n!*(1-n*x)^(r+1)),n=0..infinity). - _Vladeta Jovovic_, Feb 05 2008
%F Let A be the upper Hessenberg matrix of order n defined by: A[i,i-1]=-1, A[i,j]=binomial(j-1,i-1), (i<=j), and A[i,j]=0 otherwise. Then, for n>=3, a(n+1)=(-1)^(n-3)*coeff(charpoly(A,x),x^3). [_Milan Janjic_, Jul 08 2010]
%F E.g.f.: (1/3!) * Integral (x^3 * exp(exp(x) - 1)) dx. - _Ilya Gutkovskiy_, Jul 10 2020
%e a(5) = 4 because the partitions of {1,2,3,4,5} with 3 pairs of consecutive integers are 1234/5,123/45,12/345,1/2345.
%p seq(binomial(n-1,3)*combinat[bell](n-4),n=4..25);
%Y Cf. A105479, A105481, A105485, A105490.
%K easy,nonn
%O 4,2
%A _Augustine O. Munagi_, Apr 10 2005