OFFSET
0,1
COMMENTS
By using the approximation formula k! = (k/e)^k one can show that a(n) will be approximately 7.56*10^n.
FORMULA
a(n) = Min_{k: (k+1)/2 - (k!)^(1/k) > 10^n}.
EXAMPLE
(80+1)/2 - (80!)^(1/80) = 9.9026... < 10^1 < 10.032... = (81+1)/2 - (81!)^(1/81)
So 81 is the smallest k where the required difference exceeds 10, thus a(1) = 81.
PROG
(PARI) f(n)=return(log(sqrt(2*Pi))+(n+0.5)*log(n)-n+1/(12*n)) for(k=0, 24, n=0; forstep(i=4*k+8, 0, -1, m=n+2^i; \ if(f(m)>m*log((m+1)/2-10^k), n=m)); print1(n+1, ", ")) \\ Robert Gerbicz, Aug 24 2006
CROSSREFS
KEYWORD
nonn
AUTHOR
Stefan Steinerberger, Nov 05 2005
EXTENSIONS
More terms from Robert Gerbicz, Aug 24 2006
STATUS
approved