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^...
Terry Tao's user avatar
  • 108k
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 ...
Henri Cohen's user avatar
  • 11.5k
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 ...
Terry Tao's user avatar
  • 108k
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 ...
Christian Remling's user avatar
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}^\...
Ira Gessel's user avatar
  • 16.2k
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 ...
Carlo Beenakker's user avatar
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 ...
Ash Malyshev's user avatar
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 ...
Henri Cohen's user avatar
  • 11.5k
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 ...
Iosif Pinelis's user avatar
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-...
GH from MO's user avatar
  • 98.2k
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{...
Peter Taylor's user avatar
  • 6,516
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.
user35593's user avatar
  • 2,286
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 ...
Notamathematician's user avatar
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 ...
Zach Hunter's user avatar
  • 3,393
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 ...
Ira Gessel's user avatar
  • 16.2k
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) ...
Ira Gessel's user avatar
  • 16.2k
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+...
Iosif Pinelis's user avatar
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 ...
ouee's user avatar
  • 31
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. ...
Martin Rubey's user avatar
  • 5,533
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}...
Peter Mueller's user avatar
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)}}\...
xiaoshuchong's user avatar
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 ...
Ofir Gorodetsky's user avatar
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 ...
Faoler's user avatar
  • 431
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 ...
Oleksandr  Kulkov's user avatar
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 $$\...
Peter Taylor's user avatar
  • 6,516
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}&\...
G. Melfi's user avatar
  • 388
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 ...
Somos's user avatar
  • 2,464
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) \\ ...
Gerald Edgar's user avatar
  • 40.2k
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 ...
Евгений Павлов's user avatar
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 ...
AspiringMat's user avatar

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