A375081 Smallest k>n such that the denominator of Sum {i=n..k} (1/i) is larger than the denominator of Sum {i=n..k+1} (1/i).

5, 5, 5, 17, 17, 14, 14, 14, 14, 14, 32, 34, 34, 34, 27, 27, 27, 27, 23, 23, 27, 51, 51, 51, 51, 44, 44, 44, 44, 44, 39, 39, 39, 39, 39, 44, 74, 74, 74, 74, 74, 74, 74, 74, 65, 65, 65, 65, 65, 65, 65, 65, 65, 65, 59, 71, 71, 71, 71, 71, 71, 71, 71, 76, 76, 76

Offset: 1

Author

Ralf Stephan, Jul 29 2024

Keywords

• nonn
• changed

Examples

			1/3+1/4+1/5=47/60 and 1/3+1/4+1/5+1/6=19/20, and 60>20, so a(3)=5.
		

Links

• Bhavik Mehta, Table of n, a(n) for n = 1..10000
• Thomas Bloom, Problem 290, Erdős Problems.
• Wouter van Doorn, On the non-monotonicity of the denominator of generalized harmonic sums, arXiv:2411.03073 [math.NT], 2024.

Programs

PARI

a(n) = for(k=0, oo, my(s=sum(n=n, n+k, 1/n)); if(denominator(s)>denominator(s+1/(n+k+1)), return(n+k); break))

Python

from fractions import Fraction
from itertools import count
def A375081(n):
    a = Fraction((n<<1)+1,n*(n+1))
    for k in count(n+1):
        if a.denominator > (a:=a+Fraction(1,k+1)).denominator:
            return k # Chai Wah Wu, Jul 30 2024

Formula

a(n) < 4.374*n for all n > 1. - Wouter van Doorn, Nov 06 2024

a(n) > n + 0.54*log(n) for all large enough n, and there are infinitely many n with a(n) < n + 0.61*log(n). - Wouter van Doorn, Feb 06 2025

Extensions

a(56) onwards from Bhavik Mehta, Jul 31 2024