A307175 Smallest power to which 1+1/n must be raised in order for an interval [k,k+1], with k an integer, to be skipped.

4, 7, 9, 13, 15, 18, 22, 26, 27, 32, 33, 40, 42, 48, 51, 55, 58, 62, 66, 71, 75, 80, 85, 85, 91, 97, 103, 105, 111, 112, 120, 121, 129, 131, 139, 142, 143, 153, 156, 158, 168, 172, 175, 178, 181, 193, 197, 201, 206, 210, 215, 220, 225, 230, 235, 241, 246, 252

Offset: 2

Author

Alex Costea, Mar 27 2019

Keywords

• nonn

Comments

Here the skipping of an interval means that the interval falls strictly between (1+1/n)^(a(n)-1) and (1+1/n)^a(n).

The sequence is not monotonically increasing; a(24) = a(25) and a(62) > a(63) are the first counterexamples.

Asymptotic to n * log(n), and as such also to the prime numbers (A000040).

Examples

			1.1^26 = 11.918... and 1.1^27 = 13.109...; [12,13] is skipped, and this is the first time this happens, thus a(10)=27.
		

Crossrefs

Cf. A031435.

Programs

Mathematica

a[n_, m_] := Reduce[(1+1/n)^(m-1) < k < k+1 < (1+1/n)^m, k, Integers];
a[n_] := For[m = 1, True, m++, If[a[n, m] =!= False, Return[m]]];
Table[a[n], {n, 2, 100}] (* Jean-François Alcover, Jul 07 2019 *)

PARI

a(n) = my(k=2, last=1+1/n); while(floor(new = (1+1/n)^k) - ceil(last) != 1, k++; last = new); k; \\ Michel Marcus, Mar 30 2019

Python

from math import floor, log
def get_a_of_n(i):
     x=1+1/i
     j=i
     while floor(log(j, x))!=floor(log(j+1, x)):
         j+=1
     return floor(log(j, x))+1
def main():
     step=1
     i=2
     while True:
         y=get_a_of_n(i)
         print(y, end=", ")
         i+=step