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a(n) = Sum_{k=0..n} |Stirling1(n, k)|*(n-k)! for n>=0.
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%I #11 Jan 22 2018 03:20:27

%S 1,1,2,8,65,957,22512,773838,36561289,2271696241,179538792358,

%T 17584290721868,2090031277816649,296326507395472205,

%U 49400463740287289892,9566059122999739401954,2129221864475839211318769,539805407803681202368358785,154636541536285163968515043306,49702496963149041682740769491568

%N a(n) = Sum_{k=0..n} |Stirling1(n, k)|*(n-k)! for n>=0.

%C Compare to A007840(n) = Sum_{k=0..n} |Stirling1(n, k)|*k!, which equals the number of factorizations of permutations of n letters into ordered cycles.

%C For n > 1, a(n) is equal to the permanent of the (n-1) X (n-1) matrix in which the (i, j)-entry is equal to delta(i, j) + i, letting delta denote the Kronecker delta function, as illustrated in the below example. - _John M. Campbell_, Jan 21 2018

%F a(n) ~ ((n-1)!)^2. - _Vaclav Kotesovec_, Sep 30 2014

%e For example, the (5-1) X (5-1) matrix of the form indicated above is equal to

%e [2 1 1 1]

%e [2 3 2 2]

%e [3 3 4 3]

%e [4 4 4 5]

%e and the permanent of the above matrix is equal to 957 = a(5). - _John M. Campbell_, Jan 21 2018

%t Table[Sum[Abs[StirlingS1[n,k]]*(n-k)!,{k,0,n}],{n,0,20}] (* _Vaclav Kotesovec_, Sep 30 2014 *)

%o (PARI) {Stirling1(n, k)=if(n==0, 1, n!*polcoeff(binomial(x, n), k))}

%o {a(n)=sum(k=0, n, (-1)^(n-k)*Stirling1(n, k)*(n-k)!)}

%o for(n=0,20,print1(a(n),", "))

%Y Cf. A008275 (Stirling1 numbers).

%K nonn

%O 0,3

%A _Paul D. Hanna_, Sep 29 2014