|
%I #5 Mar 26 2020 09:57:21
%S 2,3,7,14,27,53,103,199,383,738,1422,2739,5274,10156,19556,37657,
%T 72511,139625,268857,517702,996870,1919539,3696200,7117277,13704787,
%U 26389471,50814668,97846995,188410842,362797502,698590519,1345182118,2590236887
%N Least k such that f(k) < 0, where f(1) = n and f(k) = f(k-1)*e^(1/k) - 1 for k > 1.
%C If f(k) = f(k-1)*c^(1/k) - 1 for k > 1 and c > e = 2.718281828..., then as long as f(1) is large enough, f(k) can always be positive.
%C It appears that lim_{n->infinity} a(n+1)/a(n) = 1.92556595....
%F a(n) <= A002387(n+1). Proof: since e < (1 + 1/(k-1))^k for any k > 1, f(k)/k < f(k-1)/(k-1) - 1/k < f(1) - Sum_{i=2..k} 1/i. If k = A002387(n+1), then Sum_{i=2..k} 1/i > n and f(k)/k < 0.
%e For n=0, f(1) = 0 and f(2) = -1 < 0. Thus, a(0) = 2.
%e For n=1, f(1) = 1, f(2) = sqrt(e) - 1 and f(3) = -0.094636534... < 0. Thus, a(1) = 3.
%o (PARI) a(n) = {my(k=1, f=n); while(f>0, f=f*exp(1/k++)-1); k+!n; }
%Y Cf. A001113 (e), A002387.
%K nonn,more
%O 0,1
%A _Jinyuan Wang_, Mar 26 2020
|
