16
votes
How many digits of $\sqrt{2}$ are known to date?
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 "...
- 22.8k
16
votes
Accepted
How many digits of $\sqrt{2}$ are known to date?
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: $$\begin{align} a(n) &= \frac{1}{2} - \...
- 1,565
12
votes
Accepted
On 12 cfracs: for Catalan's $K$, Gieseking's $\kappa$, and $\pi^2$, $\pi^3$, plus three for $\zeta(3)$ using Zagier's "six sporadic sequences"
We have $$C_2(-17,-6,-72)=-(5/8)L(\chi_{-3},2)$$ and
$$C_2(10,3,-9)=(1/2)L(\chi_{-3},2)$$ so both are proportional to what you
call Gieseking's constant but which is simply the value at 2 of the
L ...
- 10.4k
11
votes
Probability of winning game whereby $T+1$ heads in a row of a coin flip is required to win where $T$ is the number of cumulative tails flipped
The probability of not winning is
$$
\prod_{T=1}^\infty \left(1 - \frac1{2^T} \right)
= \frac12 \frac34 \frac78 \frac{15}{16} \cdots
= 0.28878809508660 \ldots ;
$$
that's a well-known constant (equal ...
- 75.3k
6
votes
Fibonacci series captures Euler $e=2.718\dots$
The identity given is simpler than it seems. Suppose
that $z$ is any nonzero complex number. Then
$$ \sum_{k=0}^\infty \frac{z^{n+k}}{k!} =
z^n \sum_{k=0}^\infty \frac{z^k}{k!} = e^z z^n, \quad
\sum_{...
- 2,359
4
votes
Probability of winning game whereby $T+1$ heads in a row of a coin flip is required to win where $T$ is the number of cumulative tails flipped
The game can be modeled by the following Markov chain with infinitely many states, the state $\bbox[yellow]{\boxed 0} $ is final, entering this state means winning. The other states are
$\boxed 1, \...
- 1,608
4
votes
Accepted
2D lattice sum with numerator
The $n=4$ sum is accessible as follows. Expand $(j+k)^4$; by symmetry
the odd terms $4j^3k$ and $4jk^3$ cancel out so we need only sum
$(j^4 + 6j^2k^2 + k^4) / (j^2+k^2)^4$. You say you already know
...
- 75.3k
4
votes
How many digits of $\sqrt{2}$ are known to date?
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 ...
- 72.7k
3
votes
On levels $6$ and $10$ of the McKay-Thompson series of the Monster
For $s_{10C}$, Maple finds this $7$-term recurrence:
...
- 39.4k
3
votes
Sequence that sums up to the number of permutations avoiding the pattern $1-23-4$
The following is not a complete answer, but too long for a comment.
It looks like FindStat provides an interpretation of the numbers $b(k)$ as a statistic on the set of $1-23-4$ avoiding permutations. ...
- 5,263
3
votes
On four Ramanujan-type "Legendrian" sequences with a 3-term recurrence?
Stupid of me. As O. Gorodetsky mentions, these are classical:
$$F_1=(91\zeta(3)-2\pi^3\sqrt{3})/432$$
$$F_2=(28\zeta(3)-\pi^3)/64$$
$$F_3=(117\zeta(3)-2\pi^3\sqrt{3})/243$$
In addition, note that ...
- 10.4k
2
votes
Accepted
convergence for a series
This is not a research level question, but I feel like answering it. Simply, use the elementary estimate
$$\log(m!)\sim m\log m.$$
The symbol $\sim$ means that the ratio of the two sides tends to $1$. ...
- 90.6k
2
votes
On 12 cfracs: for Catalan's $K$, Gieseking's $\kappa$, and $\pi^2$, $\pi^3$, plus three for $\zeta(3)$ using Zagier's "six sporadic sequences"
Zagier answered your first question in his original paper. See the table in p. 11 here. He gives there evaluations of the continued fractions associated with his sequences A, C, D, E and F (note that ...
- 11.6k
2
votes
Finding a new level-$7$ pi formula using the relation $j_{7A}(\tau) = \left(\sqrt{j_{7B}(\tau)}+\frac{7}{\sqrt{j_{7B}(\tau)}}\right)^2$
For Question $3$ about the recurrence relations, using Mathematica,
for $a_n := T_{7A}(n)$ I found:
$$ 0 = 14(n+1)(n+2)(2n+3) a_n \\
-3(n+2)(19n^2+76n+80) a_{n+1} \\
+ 5(2n+5)(3n^2+15n+19) a_{n+2} \\...
- 2,359
2
votes
Finding a new level-$7$ pi formula using the relation $j_{7A}(\tau) = \left(\sqrt{j_{7B}(\tau)}+\frac{7}{\sqrt{j_{7B}(\tau)}}\right)^2$
For question 3.
Using Maple's gfun library, I get this $4$-term recurrence for $t_{7A}$:
\begin{align}
0 &=14\, \left( 2\,n+3 \right) \left( n+2 \right) \left( n+1 \right) u
\left( n \right)
\\ ...
- 39.4k
1
vote
Transformations of Ramanujan's 1/pi formulas $\sum_{n=0}^{\infty} s(n)\frac{An+ B}{C^n}$ and Monster moonshine functions
(This is an addendum to add more details but not to clutter the main post.)
As more evidence, the method gives alternative closed-forms for the level-$6$ binomial sums. To recall, we have,
$$j_{6A} = \...
- 11.3k
1
vote
Accepted
Sequences that sum up to Dowling numbers
Cleaning up the notation a bit,
$$b_{m,k}(n) = m\, b_{m,k}(n-2^{\ell(n)}) + k \sum_{j=0}^{\ell(n)-1} [n \,\&\, 2^j = 0] \,b_{m,k}(n - 2^{\ell(n)} + 2^j)$$ where $\&$ is bitwise AND.
$$s_{m,k}(...
- 5,582
1
vote
Advanced software for OEIS?
This should be useful
https://sequencedb.net/
looks pretty sharp to me.
Sequences are longer than on OEIS.
- 1,266
Only top scored, non community-wiki answers of a minimum length are eligible