To determine if we can prove that for ( n > 8 ), both ( h_n ) and ( h_{n+1} ) cannot be prime, we start by analyzing the properties of harmonic numbers and their numerators ( h_n ).

Key Observations:

1. Harmonic Numerators (( h_n )): The ( n )-th harmonic number ( H_n = \sum_{k=1}^n \frac{1}{k} ) can be expressed as ( \frac{a}{b} ) in reduced form, where ( a = h_n ) (the...

Write Hn = 1 + 1/2 + … + 1/n = An/Bn in lowest terms, and hn = An.

A key 3‑adic fact about harmonic numbers (going back at least to work of Eswarathasan and Levine, Math. Comp. 1989) is the following.

Lemma (3‑adic divisibility). Let e = ⌊log3 n⌋. Then 3 divides the reduced numerator hn if n ≥ 2·3^e. Equivalently, among integers with the same highest power of 3 not exceeding them (i.e., n ∈ [3^e...

Write

    H_n = 1 + 1/2 + … + 1/n = a_n / b_n   (a_n , b_n)=1 ,   a_n = h_n .

A prime value of a_n is sometimes called a “harmonic–number prime”.
A great deal is known about the divisibility of the numerators a_n (see, for example, Eswarathasan & Levine, J. Number Th. 23 (1986), 153-160, and the survey by Brent & Cohen in “Number Theory and Cryptography”, 1990), but at present we do **n...

Write the n-th (ordinary) harmonic number in lowest terms as

            Hₙ = 1 + ½ + ⅓ + … + 1/n = Aₙ / Bₙ ,      gcd(Aₙ , Bₙ)=1 .  

The integers Aₙ are collected in OEIS A002805 and are sometimes called the “harmonic numerators”.
So far the values of n for which Aₙ is a (decimal) prime have been computed far beyond 10 000; the list that you quote agrees with what is presently known.

W...

Answering this question involves a fascinating piece of number theory related to harmonic numbers, specifically Wolstenholme's Theorem. While proving that there are finitely or infinitely many such primes is an open and likely very difficult problem, we can make significant headway on the question about consecutive prime numerators.

Let's break down the proof for why it's conjectured that for $n ...

First, recall that the nth harmonic nu...

Recall that if we write the th harmonic number in lowest terms [ H_n ;=;\sum_{k=1}^n\frac1k ;=;\frac{A_n}{B_n},\qquad\gcd(A_n,B_n)=1, ] then one studies when the numerator (A_n) (often denoted (h_n)) is prime. Numerically one finds [ n=2,3,5,8,9,21,26,41,56,62,69,79,89,91,122,\dots ] and among those the only consecutive pair is (n=2,3) and the next is (n=8,9). The question i...

Looking at this problem, I need to understand what's being asked. The harmonic numbers are: where , and we define .

The question asks whether both and can be prime for .

Let me first verify the claim by checking some values. From the given list, we have:

• is prime
• is prime

S...

Analyzing Consecutive Prime Numerators in Harmonic Numbers

The question asks if we can prove that for , both and cannot be prime simultaneously, where represents the numerator of the -th harmonic number when expressed as a reduced fraction.

Establishing the Relationship Between and

Let's start by examining how relates to :

Whe...