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Let $b(n)$ denote the Euler quotient modulo $n$.

In OEIS we have A128465 Numbers k such that k divides the numerator of alternating Harmonic number H'((k+1)/2)

For $n>1$ we have $b(A128465(n))=2$.

Also all primes up to $10^8$ with $b(n)=2$ are in the sequence.

Q1 Is this relation between the sequence and Euler quotients true?

There are too few terms for experimental evidence.

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  • $\begingroup$ The "Euler quotient" $b(n)$ is $(2^{\phi(n)}-1)/n$, reduced modulo $n$. The alternating harmonic number $H'(m)$ is $\sum_1^m(-1)^{k+1}/k$. $\endgroup$
    – Gerry Myerson
    Commented Nov 30, 2022 at 1:37
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    $\begingroup$ Would you be willing to change "Euler quotient modulo $p$ two" to "Euler quotient $2$ modulo $p$"? It took me forever to parse the title, and I'm still not sure I got it right. $\endgroup$
    – LSpice
    Commented Nov 30, 2022 at 16:03

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While no composite terms of A128465 are known, here is a proof that an odd prime $p$ belongs to A128465 if and only if $b(p)\equiv 2(-1)^{\tfrac{p+1}2}\pmod{p}$.

First notice that for an odd prime $p$, $b(p)$ equals Fermat quotient $q_p(2)$ modulo $p$.

Then, $$H'(\tfrac{p+1}2)=\sum_{k=1}^{\tfrac{p+1}2} \frac{(-1)^{k-1}}{k}\equiv 2(-1)^\tfrac{p-1}2-\frac12 H(\tfrac{p-1}2)\pmod{p}.$$

It follows that an odd prime $p$ belongs to A128465 if and only if $$H(\tfrac{p-1}2) \equiv 4(-1)^\tfrac{p-1}2 \pmod{p}.$$ On the other hand, the formula (45) in Lehmer (1938) states: $$H(\tfrac{p-1}2) \equiv -2q_p(2) + pq_p(2)^2\pmod{p^2}.$$

That is, an odd prime $p$ belongs to A128465 if and only if $$q_p(2)\equiv 2(-1)^\tfrac{p+1}2 \pmod{p}.$$ QED

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  • $\begingroup$ Thanks. Could there be generalization for divisibility and fixed Fermat quotient? $\endgroup$
    – joro
    Commented Dec 1, 2022 at 8:28
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    $\begingroup$ @joro: Some generalizations are given in the cited Lehmer's paper. $\endgroup$
    – Max Alekseyev
    Commented Dec 1, 2022 at 11:53

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