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 A080073 Let f(n)! = n^n. Then f(n) = n g(1/log(n)), where g has the asymptotic series g(x) = Sum a(j) x^j/j!. The given sequence is a(j). 0

%I

%S 1,1,0,-3,4,50,-264,-1638,25264,40896,-3357360,13380840,559239264,

%T -7126367664,-98536058880,3137828374800,8293939695360,

%U -1427422903584000,10789876955529216,666226173751955712,-14427332604300810240,-279534553922071445760

%N Let f(n)! = n^n. Then f(n) = n g(1/log(n)), where g has the asymptotic series g(x) = Sum a(j) x^j/j!. The given sequence is a(j).

%F E.g.f. A(x), safisfies A(x)=1+x*(A(x))*(1-log(A(x)),

%F a(n)=((n-1)!*sum(i=0..n-1, (binomial(n,i)*sum(j=0..n, j!*(-1)^(j)*binomial(n,j)*stirling1(n-i-1,j)))/(n-i-1)!)), n>0, a(0)=1. [_Vladimir Kruchinin_, Oct 13 2012]

%e f(n) = n (1 + 1/log(n) - 1/(2 log(n)^3) + ...), so a(0) = 1, a(1) = 1, a(2) = 0 and a(3) = (-1/2)*3! = -3.

%o (Maxima)

%o a(n):=if n=0 then 1 else ((n-1)!*sum((binomial(n,i)*sum(j!*(-1)^(j)*binomial(n,j)*stirling1(n-i-1,j),j,0,n))/(n-i-1)!,i,0,n-1)); [_Vladimir Kruchinin_, Oct 13 2012]

%K easy,sign

%O 0,4

%A Jim Ferry (jferry(AT)alum.mit.edu), Mar 14 2003

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