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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.

%I A307175 #34 Oct 18 2024 17:59:04
%S A307175 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 A307175 91,97,103,105,111,112,120,121,129,131,139,142,143,153,156,158,168,
%U A307175 172,175,178,181,193,197,201,206,210,215,220,225,230,235,241,246,252
%N 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.
%C A307175 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 A307175 The sequence is not monotonically increasing; a(24) = a(25) and a(62) > a(63) are the first counterexamples.
%C A307175 Asymptotic to n * log(n), and as such also to the prime numbers (A000040).
%e A307175 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 A307175 a[n_, m_] := Reduce[(1+1/n)^(m-1) < k < k+1 < (1+1/n)^m, k, Integers];
%t A307175 a[n_] := For[m = 1, True, m++, If[a[n, m] =!= False, Return[m]]];
%t A307175 Table[a[n], {n, 2, 100}] (* _Jean-François Alcover_, Jul 07 2019 *)
%o A307175 (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 A307175 (Python)
%o A307175 from math import floor, log
%o A307175 def get_a_of_n(i):
%o A307175      x=1+1/i
%o A307175      j=i
%o A307175      while floor(log(j, x))!=floor(log(j+1, x)):
%o A307175          j+=1
%o A307175      return floor(log(j, x))+1
%o A307175 def main():
%o A307175      step=1
%o A307175      i=2
%o A307175      while True:
%o A307175          y=get_a_of_n(i)
%o A307175          print(y, end=", ")
%o A307175          i+=step
%Y A307175 Cf. A031435.
%K A307175 nonn
%O A307175 2,1
%A A307175 _Alex Costea_, Mar 27 2019