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Background

This question was inspired by this code golf post, and I've taken some of this background explanation from there.

Consider the generalised harmonic numbers of order 2:

$$H_{n,2} = \sum^n_{k=1} \frac 1 {k^2}$$

This sequence begins:

$$1, \frac 5 4, \frac {49} {36}, \frac {205} {144}, \dots\ \text{and converges to } \frac {\pi^2} 6\ \text{as } n \to \infty$$

However, the numerators of this sequence form another sequence known as the Wolstenholme numbers (A007406):

$$1, 5, 49, 205, 5269, 5369, 266681, 1077749, 9778141, ...$$

Question

Let $i$ be the index and $w(i)$ the Wolstenholme number at index $i$.

So $w(2) = 5$, $w(3) = 49$, and so on.

Is it the case that $w(i)$ is always greater than or equal to $i$?

The initial terms suggest the answer is yes, but it might be possible that, when reduced to lowest terms, the fraction producing $w(i)$ collapses down so much that its numerator is less than its index $i$.

Is there an argument to demonstrate that this cannot happen?

Luis Mendo
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Jonah
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1 Answers1

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One can adapt the proof that the usual harmonic numbers are never integers to show this. Suppose that $k$ is chosen so that $2^k \le n < 2^{k+1}$. Then $2^{2k}$ divides exactly one denominator in the definition of $H_{n,2}$, which implies that $2^{2k}$ divides the denominator of $H_{n,2}$ itself even when reduced to lowest terms (this implication holds for any prime power). Therefore the numerator of $H_{n,2}$ is also at least $2^{2k} > 2^{k-1}n \ge n$ when $k\ge2$, that is, when $n\ge 2^2 = 4$; and the claim can be checked by hand for $1\le n\le 3$.

Greg Martin
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