
COMMENTS

Rewriting the expression as log(m!)/m > n yields the interpretation: the smallest m such that the mean of the logs of the integers from 1 to m is greater than the integer n. This is also a useful way to determine a(n) for large n.
a(n) diverges from floor[e^(n+1)] slowly and almost linearly (to the low side) as n increases with these differences: 0, 1, 2, 2, 3, 3, 3, 3, 5, 5, 6, 6, 7, 7, 8, 8, 8, 9, 9, 10, 10.
The first five terms are the same as A128104(n1).
From Jon E. Schoenfield, Dec 07 2014: (Start)
Let N = n+1, b = log(2*Pi)/2  1/4, and u = N/2 + b; then
a(n) = ceiling(x)
where x = exp(N  (u + 1/4) / e^N  (u^2 + 1/48) / e^(2*N))
is correct for all n in [0..10000], and probably for all nonnegative n. Rationale: let t be the real number that satisfies gamma(t+1) = exp(n*t); then the smallest integer m such that m! > exp(n*m) is ceiling(t). The above formula for x gives an approximation for t. At large values of n, x becomes huge, yet the approximation error xt becomes extremely small, very roughly on the order of x^2. E.g., at n=10000, x = 2.393...*10^4347, and xt = 3.273...*10^8676, so x agrees with t to more than 13000 significant digits. The above formula for a(n) will fail if and only if x and t fall on opposite sides of an integer, which seems extremely unlikely ever to occur. (End)
