How many digits of $\sqrt{2}$ are known to date?

Question

How many digits of $\sqrt{2}$ are known to date, in base 10 and in base 2? I am trying to produce the largest sequence known to date, and would like to sense if I can do it either alone or with hiring someone. I should have no problems producing 1 trillion, but almost sure I can't produce 1 quadrillion in any reasonable amount of time.

The goal is for marketing purposes and not to impress the mathematical community, but instead to attract clients to buy services that I offer. I'd like to also know about the digits of $\pi$, since producing a longer list (albeit for $\sqrt{2}$ or the golden ratio) would have a stronger impact.

Update on 5/4/2023

I am aware of what is in Wikipedia on this subject, and about the integer square root and how to compute it efficiently (for instance using the gmpy2 library). I develop PRNGs based on digits of millions of quadratic irrationals using new fast formulas, starting at arbitrary large locations, see here. If I claim that I can do better than what is in Wikipedia because I don't know the most recent computations, I could be accused of false advertising when stating that I have the longest sequence.

I want to avoid this, thus my question. The target customers will understand the random character of these digits, especially if offering a competition featuring 50k previous digits of one of these numbers starting at some location, and offer a large award (that I know no one will win) for correctly predicting the next 20k digits (with participants not knowing which starting location and which quadratic irrational I use).

I've made quite a bit of research on this topic, for instance - among others - proving that the digits of $\sqrt{2}$ and $\sqrt{3}$ are uncorrelated. The definition of correlation is in the comments.

Answer 1

In one sense, all of the base-2 digits (or, I guess, "bits") of $\sqrt{2}$ are known because we have a closed-form formula, according to the OEIS: $$\begin{align} a(n) &= \frac{1}{2} - \frac{2\arctan(\cot(2^{-\frac{3}{2}+n}\pi))}{\pi} + \frac{\arctan(\cot(2^{-\frac{1}{2}+n}\cdot\pi))}{\pi} \\[10pt] &= \left\lfloor 2^{-\frac{1}{2}+n}\right\rfloor -2\left\lfloor 2^{-\frac{3}{2}+n}\right\rfloor \end{align}$$

ETA:

According to Wikipedia, the record number of digits for $\sqrt{2}$ as of this writing is $10^{13}+1000$. Further, this website purports to contain information regarding world records for the computation of various constants. I am unsure of how credible that site is but it came up in the citations on the Wikipedia page.

Answer 2

My (very limited) understanding of the paper

Polynomial Factorization and Nonrandomness of Bits of Algebraic and Some Transcendental Numbers

is that bit strings of algebraic numbers are not "cryptographically secure". The authors (Kannan, Lenstra and Lovász) write as the first sentence of the abstract:

We show that the binary expansions of algebraic numbers do not form secure pseudorandom sequences

Edit: Here is a bit of their first paragraph:

We show that if a complex number $a$ satisfies an irreducible polynomial $h(X)$ of degree $d$ with integral coefficients in absolute value at most $H$, then given $O(d^2 + d \cdot \log H)$ bits of the binary expansion of the real and complex parts of $a$, we can find $h(X)$ in deterministic polynomial time (and then compute in polynomial time any further bits of $a$).

So in your case $a = 2^n \sqrt{q} - p$ solves the polynomial equation

$$(X + p)^2 - 2^{2n}q = X^2 + 2pX + (p^2 - 2^{2n}q) = 0.$$

This has degree $d = 2$. Both $p$ and $2^{2n}$ have bit-size $O(n)$. If $q$ also has bit-size $O(n)$, and if you provide $O(d^2 + d \log H) = O(4 + 2 n) = O(n)$ bits, then their attack applies.

---

I'll end with the canonical (and very flashy but perhaps slightly unfair!) quote from von Neumann, found in his paper Various techniques used in connection with random digits:

Anyone who considers arithmetical methods of producing random digits is, of course, in a state of sin. For, as has been pointed out several times, there is no such thing as a random number --- there are only methods to produce random numbers, and a strict arithmetic procedure of course is not such a method.

Answer 3

This is a comment about your proposed application to PRNGs. If you're trying to "sell" a PRNG, then most customers will want to know if your method is faster and more secure than some standard cryptographic RNG, such as one based on the Advanced Encryption Standard. While I obviously don't know the details of your proposed method, I suspect that it will not be faster or more secure than the Advanced Encryption Standard. In any case, that's the sort of "baseline" that you should be using for comparison, if PRNGs really are your main "product."