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A145913
a(n) = smallest integer k such that 1/log((1 + k)^(1/k)) is bigger than n.
2
1, 3, 6, 10, 14, 18, 22, 27, 32, 37, 42, 47, 52, 57, 63, 68, 74, 79, 85, 91, 97, 102, 108, 114, 120, 126, 133, 139, 145, 151, 157, 164, 170, 176, 183, 189, 196, 202, 209, 216, 222, 229, 235, 242, 249, 256, 262, 269, 276, 283, 290, 297, 304, 310, 317, 324, 331
OFFSET
1,2
COMMENTS
Note that 1/log((1 + k)^(1/k)) = 1/Hypergeometric2F1(1,1,2,-z).
Smallest positive integer k such that k > n*log(1 + k). - Peter Munn, Mar 21 2017
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 1000 terms from Indranil Ghosh)
FORMULA
a(n) ~ n log n. - Charles R Greathouse IV, Mar 24 2017
MATHEMATICA
a = {}; k = 1; Do[If[N[((1/n) Log[1 + n])^(-1)] > k, AppendTo[a, n]; k = k + 1], {n, 1, 1010}]; a
PROG
(PARI) a(n) = {my(k=1); while(1, if((1/log((1 + k)^(1/k))) > n, return (k), k++)); };
for(n=1, 100, print1(a(n), ", ")) \\ Indranil Ghosh, Mar 23 2017
(PARI) a(n)=my(k=solve(x=n*log(n), n^2, n*log(x+1)-x)\1); while(k <= n*log(k+1), k++); k \\ Charles R Greathouse IV, Mar 24 2017
(Python)
import math
def a(n):
k=1
while True:
if (1/math.log((1 + k)**(1/k))) > n: return k
else: k+=1
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Mar 23 2017
CROSSREFS
Cf. A145914.
Sequence in context: A338074 A310064 A113127 * A130246 A167381 A269745
KEYWORD
nonn
AUTHOR
Artur Jasinski, Oct 24 2008
STATUS
approved