# Andre LeClair, Riemann zeta zero approximation?

This sequence A177885 in the oeis seemingly relates imaginary parts of non-trivial Riemann zeta zeros with the LambertW function. The real and imaginary parts of the Riemann zeta function is the sum of cosine and sine waves with logarithms as frequencies.

Logarithms can be calculated as:

$$\log(n)=\lim_{s\to 1} \, \left(1-\frac{1}{n^{s-1}}\right) \zeta (s)$$

of which the numerators in the Dirichlet series are found in the following infinite table:

$$T = \begin{bmatrix} 0&0&0&0&0&0&0 \\ 1&-1&1&-1&1&-1&1 \\ 1&1&-2&1&1&-2&1 \\ 1&1&1&-3&1&1&1 \\ 1&1&1&1&-4&1&1 \\ 1&1&1&1&1&-5&1 \\ 1&1&1&1&1&1&-6 \end{bmatrix}$$

which has the definition:

$$T(n,k) = -(n-1)\; \text{ if }\; n|k, \;\text{ else } \;1,$$

Repeating/recursing the formula above we write:

$$\log(a(n))= \lim_{s\to 1} \, \zeta (s) \sum _{k=1}^n \frac{T(n,k)}{k^{s-1}}$$

where a(n) appears to be: $$a(n)=\frac{n^n}{n!}$$

$a(n) =$ {1, 2, 9/2, 32/3, 625/24, 324/5, 117649/720, 131072/315, 4782969/4480, 1562500/567, 25937424601/3628800, 35831808/1925,...}

$\left\{1,2,\frac{9}{2},\frac{32}{3},\frac{625}{24},\frac{324}{5},\frac{117649}{720},\frac{131072}{315},\frac{4782969}{4480},\frac{1562500}{567},\frac{25937424601}{3628800},\frac{35831808}{1925}\right\}$

multiplying with the factorial one finds the similar but alternating sequence A177885 in the oeis.

There in the comment this approximate formula is given:

Table[N[1/2 + 2*Pi*Exp[1]*(n - 11/8)/Exp[1]/LambertW[(n - 11/8)/Exp[1]]*I], {n,
1, 12}]
Table[N[ZetaZero[n]], {n, 1, 12}]


which gives:

{0.5 + 14.5213 I, 0.5 + 20.6557 I, 0.5 + 25.4927 I, 0.5 + 29.7394 I,
0.5 + 33.6245 I, 0.5 + 37.2574 I, 0.5 + 40.7006 I, 0.5 + 43.994 I,
0.5 + 47.1651 I, 0.5 + 50.2337 I, 0.5 + 53.2144 I, 0.5 + 56.1189 I}

{0.5 + 14.1347 I, 0.5 + 21.022 I, 0.5 + 25.0109 I, 0.5 + 30.4249 I,
0.5 + 32.9351 I, 0.5 + 37.5862 I, 0.5 + 40.9187 I, 0.5 + 43.3271 I,
0.5 + 48.0052 I, 0.5 + 49.7738 I, 0.5 + 52.9703 I, 0.5 + 56.4462 I}


The Series for x/LambertW is:

Series[x/LambertW[x], {x, 0, 7}]


$$\frac{x}{W(x)} = 1+x-\frac{x^2}{2}+\frac{2 x^3}{3}-\frac{9 x^4}{8}+\frac{32 x^5}{15}-\frac{625 x^6}{144}+\frac{324 x^7}{35}+O\left(x^8\right)$$

which has some similarity with $a(n)$

$$\frac{x}{W(x)} = \frac{(-1)^n n^n x^{n+1}}{(n+1)!}$$

$$a(n)=\frac{n^n}{n!}$$

$\left\{\frac{1}{2},\frac{2}{3},\frac{9}{8},\frac{32}{15},\frac{625}{144},\frac{324}{35},\frac{117649}{5760},\frac{131072}{2835},\frac{4782969}{44800},\frac{1562500}{6237},\frac{25937424601}{43545600},\frac{35831808}{25025}\right\}$

Is there a connection?

Edit 7.9.2013:

Would these sequences give more accurate power series approximations? Just a thought.

Clear[t, s, nn, m, k, n];
m = 1;
nn = 12;
t[n_, 1] = 1;
t[1, k_] = 1;
t[n_, k_] := t[n, k] = (1 - If[Mod[k, n] == 0, n, 0]);
MatrixForm[Table[Table[t[n, k], {k, 1, m*nn}], {n, 1, m*nn}]];
Print["here"]
Monitor[A =
Table[Limit[Zeta[s]*Sum[t[n, k]/k^(s - 1), {k, 1, m*n}],
s -> 1], {n, 1, nn}], n]

Clear[t, s, nn, m, k, n];
m = 2;
nn = 12;
t[n_, 1] = 1;
t[1, k_] = 1;
t[n_, k_] := t[n, k] = (1 - If[Mod[k, n] == 0, n, 0]);
MatrixForm[Table[Table[t[n, k], {k, 1, m*nn}], {n, 1, m*nn}]];
Print["here"]
Monitor[A =
Table[Limit[Zeta[s]*Sum[t[n, k]/k^(s - 1), {k, 1, m*n}],
s -> 1], {n, 1, nn}], n]

Clear[t, s, nn, m, k, n];
m = 3;
nn = 12;
t[n_, 1] = 1;
t[1, k_] = 1;
t[n_, k_] := t[n, k] = (1 - If[Mod[k, n] == 0, n, 0]);
MatrixForm[Table[Table[t[n, k], {k, 1, m*nn}], {n, 1, m*nn}]];
Print["here"]
Monitor[A =
Table[Limit[Zeta[s]*Sum[t[n, k]/k^(s - 1), {k, 1, m*n}],
s -> 1], {n, 1, nn}], n]

Clear[t, s, nn, m, k, n];
m = 4;
nn = 12;
t[n_, 1] = 1;
t[1, k_] = 1;
t[n_, k_] := t[n, k] = (1 - If[Mod[k, n] == 0, n, 0]);
MatrixForm[Table[Table[t[n, k], {k, 1, m*nn}], {n, 1, m*nn}]];
Print["here"]
Monitor[A =
Table[Limit[Zeta[s]*Sum[t[n, k]/k^(s - 1), {k, 1, m*n}],
s -> 1], {n, 1, nn}], n]

Clear[t, s, nn, m, k, n];
m = 5;
nn = 12;
t[n_, 1] = 1;
t[1, k_] = 1;
t[n_, k_] := t[n, k] = (1 - If[Mod[k, n] == 0, n, 0]);
MatrixForm[Table[Table[t[n, k], {k, 1, m*nn}], {n, 1, m*nn}]];
Print["here"]
Monitor[A =
Table[Limit[Zeta[s]*Sum[t[n, k]/k^(s - 1), {k, 1, m*n}],
s -> 1], {n, 1, nn}], n]


Edit 7.9.2013:

The connection I was looking for:

$$\sum _{n=1}^{\infty} \frac{x (-x)^n \exp \left(\lim_{s\to 1} \, \zeta (s) \sum _{k=1}^n \frac{1-\text{If}[k \bmod n=0,n,0]}{k^{s-1}}\right)}{n+1}+x+1 =1+x-\frac{x^2}{2}+\frac{2 x^3}{3}-\frac{9 x^4}{8}+\frac{32 x^5}{15}-\frac{625 x^6}{144}+\frac{324 x^7}{35}-\frac{117649 x^8}{5760}+\frac{131072 x^9}{2835}-\frac{4782969 x^{10}}{44800}+\frac{1562500 x^{11}}{6237}-\frac{25937424601 x^{12}}{43545600}+\frac{35831808 x^{13}}{25025}-...$$

1 + x + Sum[
x*(-x)^n*Exp[
Limit[Zeta[s]*
Sum[(1 - If[Mod[k, n] == 0, n, 0])/k^(s - 1), {k, 1, n}],
s -> 1]]/(n + 1), {n, 1, 12}]
Series[x/LambertW[x], {x, 0, 12}]


Edit 2.10.2013: Integration is better:

Clear[x, n, k, s, a1, nn, b1]
b1 = Expand[
Sum[Exp[Limit[
1/(s - 1)*
Sum[(1 - If[Mod[k, n] == 0, n, 0])/(k)^(s - 1), {k, 1, 4*n}],
s -> 1]]*(-x)^n, {n, 0, 32}]];
a1 = 1 + Integrate[b1, x];
x = N[(1 - 11/8)/Exp[1], 30];
Print["here"]
N[2*Pi*Exp[1]*a1, 30]
N[2*Pi*Exp[1]*x/LambertW[x], 30]

Clear[x, n, k, s, a1, nn]
a1 = 1 + Integrate[b1, x];
x = N[(2 - 11/8)/Exp[1], 30];
Print["here"]
N[2*Pi*Exp[1]*a1, 30]
N[2*Pi*Exp[1]*x/LambertW[x], 30]


where the number $4$ within: {k, 1, 4*n}], can be varied for truncating the Dirichlet series for the logarithm of $n$. At least as long as the truncated Dirichlet series does not get longer than the power series, there is tendency for the Zeta zero approximations to stay close to the zeta zeros.

12.10.2013: Better integration:

Clear[x, n, k, s, a1, nn, b1]
b1 = Expand[
Sum[Exp[Limit[
Zeta[s]*Sum[(1 - If[Mod[k, n] == 0, n, 0])/k^(s - 1), {k, 1,
n}], s -> 1]]*(-x)^n, {n, 1, 32}]];
a1 = 1 + Integrate[1 + b1, x];
x = N[(1 - 11/8)/Exp[1], 30];
Print["here"]
N[2*Pi*Exp[1]*a1, 30]
N[2*Pi*Exp[1]*x/LambertW[x], 30]

Clear[x, n, k, s, a1, nn]
a1 = 1 + Integrate[1 + b1, x];
x = N[(2 - 11/8)/Exp[1], 30];
Print["here"]
N[2*Pi*Exp[1]*a1, 30]
N[2*Pi*Exp[1]*x/LambertW[x], 30]


This Excel Spreadsheet formula uses Andre LeClaire's formula to approximate the Riemann zeta zeros:

(European dot-comma)

you need to divide the result with: /2/PI()/EXP(1) and take the reciprocal.

tetration this is.

Yes of course, the smooth part of the zeros is given by

$$N(T)= \frac{T}{2\pi}\log\left(\frac{T}{2\pi e}\right)$$

This function can be inverted and is equal to $$T= \frac{2 \pi n}{W(ne^{-1})}$$

This is the main reason why the approximation works fine.

It is interesting to read these comments. The last comment provides a reason why it works fine, but the argument is erroneous, since N(T) has only been proven on the whole critical strip. Thus, one cannot derive the Lambert approximation to the zeros on the line from it, since it was never proven to be valid on the line. Unless of course one assumes the Riemann hypothesis is true.

There has been a follow-up paper with my post-doc Franca that develops these ideas much further, and provides both approximate and exact corrections to this approximation based on the Lambert function. It turns out that these corrections are necessary to capture the GUE statistics of Montgomery's and Odlyzko's conjectures. The paper is here: http://arxiv.org/abs/1307.8395

There is another paper that extends all these formulas to all Dirichlet L-functions.

-André LeClair

• With these corrections, we are able to compute Riemann zeros to 500 digit accuracy or more in a very simple manner, see the above paper. -Andre' – André LeClair Sep 13 '13 at 23:26
• Dear Prof. LeClair, I am but a struggling self-studier deeply interested in the Riemann Zeta Function. I just took a look at the linked abstract. Am I reading it correctly that there is an indication that you may have proved the RH? Thanks, best wishes, – user12802 Sep 13 '13 at 23:40
• Are you familiar with mathoverflow.net? For professionals. There are, of course, some outstanding participants here as well. – user12802 Sep 14 '13 at 1:39
• Andrew, I am not familiar with that site. But if you think it is worth the effort, let me know how to access it. – André LeClair Sep 14 '13 at 4:48
• Just use the link or google it. Then sign up as a user, it's comparable to here. I only read posts and comments there, so I haven't actually gone through the process of participating, but I am sure you would be most welcome there and receive extremely useful replies. With regards, – user12802 Sep 14 '13 at 10:48