A307175 - OEIS

%I #34 Oct 18 2024 17:59:04

%S 4,7,9,13,15,18,22,26,27,32,33,40,42,48,51,55,58,62,66,71,75,80,85,85,

%T 91,97,103,105,111,112,120,121,129,131,139,142,143,153,156,158,168,

%U 172,175,178,181,193,197,201,206,210,215,220,225,230,235,241,246,252

%N 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.

%C 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).

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

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

%e 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.

%t a[n_, m_] := Reduce[(1+1/n)^(m-1) < k < k+1 < (1+1/n)^m, k, Integers];

%t a[n_] := For[m = 1, True, m++, If[a[n, m] =!= False, Return[m]]];

%t Table[a[n], {n, 2, 100}] (* _Jean-François Alcover_, Jul 07 2019 *)

%o (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

%o (Python)

%o from math import floor, log

%o def get_a_of_n(i):

%o      x=1+1/i

%o      j=i

%o      while floor(log(j, x))!=floor(log(j+1, x)):

%o          j+=1

%o      return floor(log(j, x))+1

%o def main():

%o      step=1

%o      i=2

%o      while True:

%o          y=get_a_of_n(i)

%o          print(y, end=", ")

%o          i+=step

%Y Cf. A031435.

%K nonn

%O 2,1

%A _Alex Costea_, Mar 27 2019