Riemann Rave III

The aforementioned moment series (for the unitary ensemble) of Keating and Snaith is defined in their paper as

$f(n) = \prod_{j=0}^{n – 1} \frac{j!}{(j + n)!}$

for which one finds the first few terms

$1$, $\frac{2}{4!}$, $\frac{42}{9!}$, $\frac{24024}{16!}$, …

It is fun to find alternative expressions with fewer factorials because the size of terms is then more immediately apparent. For example,

$f(n) = \frac{n^{n – 1}}{(2n – 1)!} \prod_{j=0}^{n-2} (n^{2} – j^{2})^{j-n+1}$

for which one can check, say the third term

$f(3) = \frac{3.3}{} \frac{1}{9.9} \frac{1}{8} = \frac{6.7}{9!} = \frac{42}{9!}$

This shows up the dependence on the factorial of squares $n^2$, which is not so apparent in the original expression. Note that $24024=$ is similarly made of pieces that are needed for a factor of $16!$ in the 4th term. The nth term provides the coefficient of the moment $\int_{0}^{T} | \zeta (0.5 + it) |^{2n} dt$ of the Riemann zeta function, as discussed by Conrey et al in the paper that appeared around the time of Keating’s Vienna talk.

It seems that single prime factors miraculously appear in the numerators of $f(n)$. Moving on to $f(5)$ we find a numerator of $$ (assuming I did the sum correctly). So if we ignore powers of 2, the first few numerators are built from single prime factors, covering mainly primes between $2n$ and $n^{2}$.


2 Responses so far »

  1. 1

    L. Riofrio said,

    These Riemann Raves are an endless source of fascination, a gift that keeps on giving. Rumours about the proof of his hypoothesis create even more intrigue.

  2. 2

    Kea said,

    Hi Louise! Yes, stranger than fiction.

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