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Number of partitions of {1...n} containing 5 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.
5

%I #13 Jul 10 2020 22:07:45

%S 1,6,42,280,1890,13104,93786,694584,5328180,42336294,348272925,

%T 2963993760,26073738236,236857536216,2219777316216,21441389281680,

%U 213260412549303,2182163481418536,22951202450444191,247914874683742728

%N Number of partitions of {1...n} containing 5 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://www.emis.de/journals/HOA/IJMMS/2005/3451.pdf">Set Partitions with Successions and Separations</a>, IJMMS 2005:3 (2005), 451-463.

%F a(n) = binomial(n-1, 5)*Bell(n-6), the case r = 5 in the general case of r pairs: c(n, r) = binomial(n-1, r)*B(n-r-1).

%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>=5, a(n+1)=(-1)^(n-5)*coeff(charpoly(A,x),x^5). [_Milan Janjic_, Jul 08 2010]

%F E.g.f.: (1/5!) * Integral (x^5 * exp(exp(x) - 1)) dx. - _Ilya Gutkovskiy_, Jul 10 2020

%e a(7) = 6 because the partitions of {1,2,3,4,5,6,7} with 5 pairs of consecutive integers are 123456/7,12345/67,1234/567,123/4567,12/34567,1/234567.

%p seq(binomial(n-1,5)*combinat[bell](n-6),n=6..26);

%Y Cf. A105481, A105487, A105491.

%K easy,nonn

%O 6,2

%A _Augustine O. Munagi_, Apr 10 2005