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A333342
Least k such that f(k) < 0, where f(1) = n and f(k) = f(k-1)*e^(1/k) - 1 for k > 1.
1
2, 3, 7, 14, 27, 53, 103, 199, 383, 738, 1422, 2739, 5274, 10156, 19556, 37657, 72511, 139625, 268857, 517702, 996870, 1919539, 3696200, 7117277, 13704787, 26389471, 50814668, 97846995, 188410842, 362797502, 698590519, 1345182118, 2590236887
OFFSET
0,1
COMMENTS
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.
It appears that lim_{n->oo} a(n+1)/a(n) = 1.92556595....
FORMULA
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.
EXAMPLE
For n=0, f(1) = 0 and f(2) = -1 < 0. So, a(0) = 2.
For n=1, f(1) = 1, f(2) = sqrt(e)-1 and f(3) = -0.094636534... < 0. So, a(1) = 3.
PROG
(PARI) a(n) = {my(k=1, f=n); while(f>0, f=f*exp(1/k++)-1); k+!n; }
CROSSREFS
Cf. A001113 (e), A002387.
Sequence in context: A054194 A138651 A131300 * A078043 A294627 A293326
KEYWORD
nonn,more
AUTHOR
Jinyuan Wang, Mar 26 2020
STATUS
approved