Sylvester's sequence - series representation of Vardi's constant

Question

First, a bit of background - Sylvester's sequence is defined recursively as $$e_{n+1}=\prod_{j=0}^{n-1}e_j=e_n^2-e_n+1, \space\space\space e_0=2$$ and, miraculously, there exists a number $E\approx 1.26$ that satisfies $$e_n=\lfloor E^{2^{n+1}}+1/2\rfloor$$ Wolfram gives a series representation for the constant $E$ (also called "Vardi's constant"), a proof of which I have been unable to find (or derive): $$E=\frac{\sqrt 6}{2}\exp\Bigg[\sum_{j=1}^\infty \frac{1}{2^{j+1}}\ln\bigg(1+\frac{1}{(2e_j-1)^2}\bigg)\Bigg]$$ or, reduced to a product, $$E=\frac{\sqrt 6}{2}\prod_{j=1}^\infty \bigg(1+\frac{1}{(2e_j-1)^2}\bigg)^{2^{-j-1}}$$ Reduced further, this is equal to $$E=\frac{\sqrt 6}{2}\prod_{j=1}^\infty \bigg(\frac{4e_{j+1}-4}{4e_{j+1}-3}\bigg)^{2^{-j-1}}$$ How might one derive this property? I haven't been able to make much progress with its derivation.

Answer 1

Caveat: this is a partial answer. I will try to explain the methodology for the construction of the generating function and the definition of the constant (it lacks detail but it should give an overview, if the reader finds errors please be free to fix them or let me know and I will try to fix them). I hope it will help as an idea of how Vardi's constant and its generating function can be developed. So below I will not include the construction of $E\approx 1.26$ that satisfies $$e_n=\lfloor E^{2^{n+1}}+1/2\rfloor$$ but a similar construction based on another constant that I will call $L$ such that $L\approx 1.597910225682488561085083110$ that satisfies $$e_n=\lfloor L^{2^{n}}+1/2\rfloor$$

The difference is that the exponent of my generating function based on the constant $L$ is $2^{n}$ instead of $2^{n+1}$, as in the case of Vardi's $E$.

1. Basically we observe that the expression below is true:

$$(e_n-1)^2 \lt e_{n+1} \lt e_n^2$$

for Sylvester's sequence:

$2, 3, 7, 43, 1807, 3263443, 10650056950807, 113423713055421844361000443, \dots$

(sequence A000058 in the OEIS).

e.g.

$$(e_0-1)^2 = (2-1)^2=1 \lt e_1=3 \lt e_0^2 = 2^2=4$$

$$(e_1-1)^2 = (3-1)^2=4 \lt e_2=7 \lt e_1^2 = 3^2=9$$

$$(e_2-1)^2 = (7-1)^2=36 \lt e_3=43 \lt e_2^2 = 7^2=49$$

$$\cdots$$

2. Following the classical Mills demonstration (1947) (please see details of the theorem and proof in the paper) we can conclude that we have a bounded monotone increasing sequence able to generate each $e_n$. The generation of the $L$ constant is as follows:

$$L_0={e_0^{\frac{1}{2^0}}}$$

$$L_1={e_1^{\frac{1}{2^1}}}$$

$$L_2={e_2^{\frac{1}{2^2}}}$$

$$\cdots$$

$$L_5={e_5^{\frac{1}{2^5}}}=1.597910225682488561085083110$$

Where $L$ will be: $$\lim_{n \to \infty} L_n \ .$$ As in the case of the classical results of Mills, we do no know if it is a rational number or whether a closed-form formula could be obtained.

So you will require more precision as $n$ grows higher. With $L_5$ we will be able to calculate accurately $\{e_0,e_1,e_2,e_3,e_4,e_5\}$

The generating function initially would be as follows (it will require a correction I will explain later):

$$e_n = \lfloor L^{2^n} \rfloor$$

e.g. (still without correction):

$$e_5 = \lfloor L^{2^5} \rfloor = 3263443$$

$$e_4 = \lfloor L^{2^4} \rfloor = \color{red}{1806}$$

$$e_3 = \lfloor L^{2^3} \rfloor = \color{red}{42}$$

$$e_2 = \lfloor L^{2^2} \rfloor = \color{red}{6}$$

$$e_1 = \lfloor L^{2^1} \rfloor = \color{red}{2}$$

$$e_0 = \lfloor L^{2^0} \rfloor = \color{red}{1}$$

As you can observe still $e_0 \dots e_4$ are not accurate, but if we add $1/2$ the correction is good enough to make the result of the floor function match the sequence (as you can see the same correction is done in the OP's generating function for Vardi's constant $E$).

$$e_n=\lfloor L^{2^{n}}+1/2\rfloor$$

And now:

$$e_5 = \lfloor L^{2^5} +1/2\rfloor = 3263443$$

$$e_4 = \lfloor L^{2^4} +1/2 \rfloor = 1807$$

$$e_3 = \lfloor L^{2^3} +1/2 \rfloor = 43$$

$$e_2 = \lfloor L^{2^2} +1/2 \rfloor = 7$$

$$e_1 = \lfloor L^{2^11/2} +1/2 \rfloor = 3$$

$$e_0 = \lfloor L^{2^0} +1/2 \rfloor = 2$$

My guess is that the correction is required because, due to the properties of the sequence, we are embedding the sequence into quadratics intervals $(e_n-1)^2 \lt e_{n+1} \lt e_n^2$ instead of intervals of the shape $e_n^2 \lt e_{n+1} \lt (e_n+1)^2$ (which is the same case as Mills' constant but for intervals based on powers of three instead of powers of two. In the case of Mills' constant the correction is not required). I know this point do require a formal explanation but I would dare to say that it is related with the definition of the quadratic intervals.

So we have been able to define the generating function:

$$e_n=\lfloor L^{2^{n}}+1/2\rfloor$$

where

$$L\approx 1.597910225682488561085083110$$

This is basically an overview and lacks a lot of details, but the point is that due to the characteristics of the sequence we can bound it into quadratic intervals. Due to this, there is a monotone increasing sequence that we can embed into a constant, such that a quadratic exponent applied to it (plus the floor function corrected by $+1/2$) is able to recover the embedded sequence. So I think it is safe to say that this has been done by Vardi as well in a similar way. About the calculations of Wolfram, I think they are done by Wolfram's engine backwards, once you have defined properly the generating function and the constant. It can be seen that the precision of $E$ depends on the iteration $e_n$ so as $n$ grows higher we need more precision of $E$ (and same for my example $L$) to recover accurately the element $e_n$ of the sequence.