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10 votes
2 answers
731 views

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"

I. Some functions As these will be used in the continued fraction evaluations below, recall the Riemann zeta function $\zeta(s),$ and Dirichlet beta function $\beta(s),$ $$\beta(s) = \sum_{n=1}^\infty\...
Tito Piezas III's user avatar
9 votes
2 answers
1k views

About a Ramanujan-Sato formula of level 10, a recurrence, and $\zeta(5)$?

I. Level 6 This is a long shot, but I am curious where it leads. Given the Dedekind eta function $\eta(\tau),$ define, $$\begin{aligned} j_{6A}(\tau) &= \Big(\sqrt{j_{6B}(\tau)} - \frac{1}{\sqrt{...
Tito Piezas III's user avatar
3 votes
1 answer
193 views

Sequences that sums up to second differences of Bell and Catalan numbers

Let $f(n)$ be A007814, the exponent of the highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$. Let $g(n)$ be A025480, $g(2n) = n$...
Notamathematician's user avatar
2 votes
1 answer
183 views

Pair of recurrence relations with $a(2n+1)=a(2f(n))$

Let $f(n)$ be A053645, distance to largest power of $2$ less than or equal to $n$; write $n$ in binary, change the first digit to zero, and convert back to decimal. Let $g(n)$ be A007814, the ...
Notamathematician's user avatar
7 votes
2 answers
265 views

On four Ramanujan-type "Legendrian" sequences with a 3-term recurrence?

I. Recurrences In a previous post, it was mentioned how Almkvist-Zudilin did a computer search for solutions to the recurrence relation, $$(n+1)^3s_{n+1}=(2n+1)(an^2+an+b)s_n+c\,n^3s_{n-1}$$ within a ...
Tito Piezas III's user avatar
7 votes
2 answers
428 views

Limit associated with complementary sequences

Define $A=(a_n)$ and $B=(b_n)$ as follows: $a_0=1$, $a_1=2$, $b_0=3$, $b_1=4$, and $$a_n=a_0b_{n-1}+a_1b_{n-2}$$ for $n \geq 2$, where $A$ and $B$ are increasing and every positive integer occurs ...
Clark Kimberling's user avatar
3 votes
2 answers
973 views

Recursive random number generator based on irrational numbers

Here $\{\cdot\}$ and $\lfloor \cdot\rfloor$ denote the fractional part and floor functions respectively. For a negative, non-integer number $x$, we use the following definition: $\{x\}=1-\{-x\}$. If $...
Vincent Granville's user avatar
5 votes
1 answer
168 views

On a generating function and vector $\nu$ of length $n$

Let $f(n)$ be an arbitrary function with integer values. Let $a(n)$ be an integer sequence such that $$ \frac{1}{1-x}=\sum\limits_{n=0}^{\infty}a(n)x^n\prod\limits_{k=1}^{n+1}(1-f(k)x) $$ Start with ...
Notamathematician's user avatar
0 votes
1 answer
101 views

Recurrence for the number of steps required to get one ball in each box

Given $n$ balls, all of which are initially in the first of $n$ numbered boxes, $a(n)$ is the number of steps required to get one ball in each box when a step consists of moving to the next box every ...
Notamathematician's user avatar
26 votes
1 answer
7k views

Elegant recursion for A301897

Let $a(n)$ be A301897, i.e., number of permutations $b$ of length $n$ that satisfy the Diaconis-Graham inequality $I_n(b) + EX_n(b) \leqslant D_n(b)$ with equality. Here $$a(n)=\frac{1}{n+1}\binom{2n}{...
Notamathematician's user avatar
6 votes
1 answer
268 views

Sequence that sums up to the number of permutations avoiding the pattern $1-23-4$

Let $a(n)$ be A113227, i.e., the number of permutations on $[n]\equiv \{1, \ldots, n\}$ avoiding the pattern $1-23-4$. The sequence begins with $$1, 1, 2, 6, 23, 105, 549, 3207, 20577, 143239, 1071704,...
Notamathematician's user avatar
3 votes
2 answers
458 views

Subsequence of the cubes

Let $p$ and $q$ be integers. Let $f(n)$ be A007814, the exponent of the highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$. Then ...
Notamathematician's user avatar
1 vote
0 answers
38 views

Sequences that sum up to $\frac{m^{n+1}}{n+1}\binom{2n}{n}{}_2F_1(1,n+\frac{1}{2}; n+2; -4m(m-1))$

Let $a(n,m)$ be an integer sequence such that $$a(n,m)=\frac{m^{n+1}}{n+1}\binom{2n}{n}{}_2F_1(1,n+\frac{1}{2}; n+2; -4m(m-1))$$ Let $$\ell(n)=\left\lfloor\log_2 n\right\rfloor$$ $$f(n)=n-2^{\ell(n)}$$...
Notamathematician's user avatar
1 vote
1 answer
153 views

Hybrid numeration system on $[0,1]^2$

Let $X_0,X_1\in [0,1]$ and $b_1,b_2>0$ be integers. We are going to create a numeration system for vectors $(X_0,X_1)$, the base being the vector $(b_1,b_2)$, as follows. Recursively define $X_k=\{...
Vincent Granville's user avatar
1 vote
1 answer
110 views

Number of steps required to get one ball in each box for $n=2^k$

Given $n$ balls, all of which are initially in the first of $n$ numbered boxes, $a(n)$ is the number of steps required to get one ball in each box when a step consists of moving to the next box every ...
Notamathematician's user avatar
1 vote
1 answer
295 views

Formula from the recurrence relation

Let $f(n)$ be A007814, the exponent of the highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$. Then we have an integer sequence ...
Notamathematician's user avatar