37
votes
Accepted
Elegant recursion for A301897
Here is an expanded version of the generating function argument I sketched in a comment.
For $i=1,2,3$, define the generating functions $F_i(x,y) := \sum_{n=0}^\infty \sum_{q=0}^\infty R(n,3q+i) x^n y^...
13
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
votes
Accepted
Remarkable recursions for the A204262
I can show the first identity $R(n,0) = f_{n+1,n+1}(0)$, as a consequence of the more general identity
$$ R(n,q) = \frac{1}{(q+1)!} f_{n+q+1,n}(n+1)\tag{1}\label{1}$$
for $n,q \geq 0$. Indeed, note ...
7
votes
Accepted
Precise asymptotic estimate of a recurrence sequence involving a square root
Your difference equation can be viewed as the Euler method for the initial value problem $y'(t)=\sqrt{y}$, $y(0)=\epsilon$, on the interval $0\le t\le 1$, with step size $h=\epsilon$.
This we can of ...
7
votes
Accepted
$R$-recursion for the A143017
We proceed by "guessing" a generating function for $R(n,q)$ and verifying that it has the right properties.
According to https://oeis.org/A143017, the generating function $G= \sum_{n=1}^\...
6
votes
About the high-order derivatives of Lambert function
Both questions are answered in the 2011 Ph.D. thesis of G.A. Kalugin, see also arXiv:1011.5940
$$\alpha_{n,k}=\sum_{m=0}^k\frac{1}{m!}{{2n-1}\choose{k-m}}\sum_{q=0}^m(-1)^q (q+n)^{m+n-1}.$$
There are ...
6
votes
How to show a function converges to 1
The urn problem is equivalent to a continuous-time process where the balls behave independently.
Balls can be either Black, White, or Gone. For each ball, there's an independent Poission process of ...
5
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 ...
5
votes
How to show a function converges to 1
The following answer is incomplete.
Clearly, $0\le f\le1$. Let $h:=1-f$, so that $0\le h\le1$, $h(a,0)=0$, and $h(0,b)=1$ if $b\ge1$. Here and in what follows, $a$ and $b$ are nonnegative integers.
We ...
4
votes
Accepted
Diophantine equations involving recurrence sequences
The authors skipped some easy steps. By the triangle inequality,
$$e^{z_1}\geq 1-|1-e^{z_1}|>1-0.95=1/20,$$
whence using also $z_1<0$,
$$e^{|z_1|}=e^{-z_1}<20.$$
Similarly,
$$e^{z_2}\geq 1-|1-...
4
votes
Accepted
Property of some permutations of non-negative integers such that $a(n)<2^k$ iff $n<2^k$
I'm going to use $\operatorname{msb}$ (for most significant bit) as an alias of $f$.
Since $q_i$ is a permutation, the property that $q_i(n)<2^k$ iff $n < 2^k$ is equivalent to $\operatorname{...
4
votes
Accepted
Finding a strictly increasing Collatz sequence of arbitrary length
Start with a number equal to $-1$ modulo $2^n$. Then, after one step, the number is $\frac{3(-1)+1}{2}=-1$ modulo $\frac{2^n}{2}=2^{n-1}$, so inductively it will increase $n$ steps.
4
votes
Accepted
How to solve recurrence relation with 2 variables?
Let
$$g(n,m)=\sum\limits_{i=0}^{n}\sum\limits_{j=0}^{m}\binom{i+j}{j}\binom{n-i+m-j}{m-j}\alpha^i\beta^j$$
I conjecture that
$$g(n,m)=\binom{n+m}{m}f(n,m)$$
Here is the PARI prog to verify this ...
4
votes
How to show a function converges to 1
Reduction:
Fix $p\in (0,1/2)$. For large $n$, it is not hard to see that $$f(1,n)>f((p-4p^2) n,n)(1-3p).$$
Indeed, consider the balls an urn's setup of Fedor Petrov. For the first $pn$ steps, we ...
3
votes
Accepted
General case of the some $R$-recursions
Let
$$
A(x,q)=\sum_{i=0}^{\infty}\frac{x^i}{\prod\limits_{j=0}^{i}(1-f(q+j)x)},
$$
so that $A(x) = A(x,0)$. Define $a(n,q)$ by
$$A(x,q) = \sum_{n=0}^\infty a(n,q) x^n,$$
so that $a(n) = a(n,0)$.
I ...
3
votes
Accepted
$R$-recursion for the A307389
Let
\begin{equation*}
A(x,q) = e^{qx}A(x)=e^{(q+1)x+(e^x-1)^2/2}
\end{equation*}
and define $a(n,q)$ by
\begin{equation*}
A(x,q) = \sum_{n=0}^\infty a(n,q) \frac{x^n}{n!}.
\end{equation*}
Then $a(n,0) ...
3
votes
Recurrence relation with two variables
The following is an answer to the original version of the question.
Your system of equations can be rewritten as follows: For integers $i$ and $j$ in $[0,n-2]$,
\begin{equation}
\begin{aligned}
(2n r+...
3
votes
How to show a function converges to 1
This can be solved using the so-called "fluid limit" for stochastic processes.
Consider $B_k, W_k$ the number of black (resp. white) balls at time $k$. Our aim is to obtain a macroscopic ...
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. ...
3
votes
Polynomial solutions to a difference equation
A few minor comments which don't fit conveniently into the comment fields (and would resist to future edits):
Assuming that a monic solution $P$ has degree $m$, then comparing coefficients at $x^{m+1}...
3
votes
About a Ramanujan-Sato formula of level 10, a recurrence, and $\zeta(5)$?
Since $\alpha_{k}$ is known, according to the below relation
\begin{eqnarray}
\sum_{k=0}^{\infty}\frac{\beta_{k}}{\left(j_{10D}(\tau)\right)^{k}}&=&{\sqrt{\frac{j_{10D}(\tau)}{j_{10A}(\tau)}}\...
3
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 ...
3
votes
Accepted
How to find the coefficient of $x^k$ in the expression $\prod_{p=1}^n (x^p+1)^p$?
I got it ...
firstly the degree of $(x^p+1)^p$ is $p^2$ So the degree of $\prod_{p=1}^n (x^p+1)^p$ is
$$N=1+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$$
now we have
$$\prod_{p=1}^n (x^p+1)^p=\sum_{p=1}^N ...
2
votes
Accepted
Formulas for partial composed product
Thanks everyone for the comments! I will try to summarize what's going on a bit, and post it as an answer, as I was not really aware of how symmetric squares and exterior squares work, so I will ...
2
votes
Solving a recurrence relation for the prime counting function?
It turns out to be more straightforward than I expected.
Let $C(z) = \sum_{i \ge 0} c_i z^i$ be the g.f. for $c_i$, excluding $c_{-1}$ since that doesn't show up in your recurrence.
Starting with $$\...
2
votes
Squares in Lucas sequences
One of the most known Lucas sequences, after the Fibonacci and the Lucas numbers, is the sequence of Pell numbers:
$P_n=\begin{cases}0&\mbox{if }n=0;\\1&\mbox{if }n=1;\\2P_{n-1}+P_{n-2}&\...
2
votes
Accepted
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 my code from
MMA question 285008
for $a_n := T_{7A}(n)$ I used
...
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)
\\ ...
2
votes
How to show a function converges to 1
I will use the probabilistic interpretation in my proof.
First, note that $P(1, n-1) = P(0,n)$ so we can reformulate the problem as follows:
"We start with a hat with $n$ black balls. When we ...
2
votes
How to find the coefficient of $x^k$ in the expression $\prod_{p=1}^n (x^p+1)^p$?
Caveat: OP asked me in the comment section how he can calculate the coefficient explicitly. This answer is mainly algorithmic (dynamic programming) and straight forward with FFT/convolutions/dynamic ...
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