All Questions
64 questions with no upvoted or accepted answers
7
votes
0
answers
456
views
On H. Cohen's four continued fractions for $\zeta(3), \zeta(5), \zeta(7)$?
After 6 years from this old MO post, I finally find in the literature polynomials of deg-$5$ for the continued fraction of $\zeta(5)$.
I. Recurrences involving $\zeta(5)$
In Cohen's 2022 paper, ...
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6
votes
0
answers
245
views
Searching for a proof of the pattern and identification of integer coefficients for the A329369
Please see the update given below. Everything you need to know from the old version of the question are the functions $a(n), \ell(n), s(n), t(n), r(n)$.
Let $a(n)$ be A329369 (i.e, number of ...
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6
votes
0
answers
257
views
Convergence with the recurrence $T_{n+1}=T_n^2-T_n+\frac{n}{p_n}$
For each integer $n\geq 1$ I define the recurrence $$T_{n+1}=T_n^2-T_n+\frac{n}{p_n},$$
with $T_1=1$, where $p_k$ denotes the $k$-th prime.
So multiplying by $(-1)^n$ and telescoping gives that for ...
4
votes
0
answers
118
views
Something (which might be called multi-continued fraction) for the A112487
Let $a(n)$ be A112487 i.e. an integer sequence with exponential generating function
$$
A(x)=\exp\left(\int (A(x)+A(x)^2)\,dx\right), \\
A(0)=1
$$
However, the definition in the name of the sequence is
...
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4
votes
0
answers
90
views
Closed form for subsequence of the partial sums of generalized A329369
Let $a(n,m,k)$ be an integer sequence such that
$$a(n,m,k)=\sum\limits_{i=0}^{n}{n\brace i}(m-k)^{n-i}\prod\limits_{j=0}^{i-1}(kj+1)$$
Here ${n\brace k}$ is the Stirling number of the second kind.
...
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4
votes
0
answers
171
views
Positivity conjecture for Somos sequences
Let $\{s_n\}$ be the Somos-$4$ sequence, which is defined by $$s_{n+4}s_n=\alpha s_{n+3}s_{n+1}+\beta s_n^2.$$ It is known that $s_n$ is a Laurent polynomial: $s_n\in\mathbb{Z}[s_1^{\pm1}, \ldots, s_4^...
- 12.3k
3
votes
0
answers
129
views
Sequence that sums up to A014307
Let $s(n,k)$ be a (signed) Stirling number of the first kind.
Let $n \brace k$ be a Stirling number of the second kind.
Let $a(n)$ be A014307. Here
$$
A(x) = \sum\limits_{k=0}^{\infty} \frac{a(k)}{...
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3
votes
0
answers
88
views
Recursion for reversed rows of the A373183 using unsigned Stirling numbers of the first kind
Let $\left[{n \atop k}\right]$ be unsigned Stirling numbers of the first kind. Here
$$
\left[{n \atop k}\right] = (n-1)\left[{n-1 \atop k}\right] + \left[{n-1 \atop k-1}\right], \\
\left[{n \atop 0}\...
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3
votes
0
answers
70
views
$R$-recursion for the A249833 (similar to A235129)
Let $a(n)$ be A249833 i.e. an integer sequence such that its exponential generating function $A(x)$ satisfies
$$
A(x) = 1 + \int A(x) + (A(x))^2\log A(x)\,dx
$$
The sequence begins with
$$
1, 1, 2, 7, ...
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3
votes
0
answers
121
views
Closed form from a slightly modified recursion for transposed Catalan triangle
Let $a_1(n)$ be A000108, i.e. Catalan numbers. Here
$$
a_1(n)=\frac{1}{n+1}\binom{2n}{n}
$$
Let $a_2(n)$ be A059715, i.e. number of multi-directed animals on the triangular lattice. From OEIS page we ...
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3
votes
0
answers
69
views
Sequence that sum up to A343685
Let $a(n)$ be A343685 i.e.
$$
a(n)=2na(n-1)+\sum\limits_{j=0}^{n-1}\binom{n}{j}(n-j-1)!a(j), \\
a(0)=1
$$
Here the exponential generating function $A(x)$ satisfy
$$
A(x)=\frac{1}{1-2x+\log(1-x)}
$$
...
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3
votes
0
answers
210
views
Using Ramanujan-type "Legendrian" sequences to find new formulas for $\frac1{\pi}$?
I. Recurrences
(Continued from this post.) In Cooper's 2012 paper, "Sporadic sequences, modular forms and new series for 1/π", he did a computer search for the recurrence relation,
$$(n+1)^3 ...
- 12.6k
3
votes
0
answers
97
views
Identical digits at the end of adjacent terms of the sequence
Let $m\geq 2$ be a fixed integer.
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 $n$.
...
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2
votes
0
answers
67
views
$R$-recursion for A006351
Let $a(n)$ be A006351 (i.e., number of series-parallel networks with n labeled edges. Also called yoke-chains by Cayley and MacMahon). Here exponential generating function is $A(x)$ such that $B(x) = ...
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2
votes
0
answers
59
views
$R$-recursion for A338193
Let $a(n)$ be A338193. Here generating function is $A(x)$ such that
$$
A(x) = 1 + \int\frac{\left(\frac{x}{A(x)}\right)'}{\left(\frac{x}{(A(x))^2}\right)'} \, dx.
$$
Let
$$
R(n, q) = \begin{cases}
1 &...
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2
votes
0
answers
46
views
On A088352 as an antidiagonal sums of A129179
Let $a(n)$ be A088352. Here $a(n)$ is an integer sequence with generating function $A(x)$ such that
$$
A(x) = \cfrac{1}{1-x-\cfrac{x^2}{1-x^3-\cfrac{x^4}{1-x^5-\cfrac{x^6}{1-x^7-\cfrac{x^8}{\ddots}}}}}...
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2
votes
0
answers
100
views
Another (unique) algorithm for the A329369
Let $a(n)$ be A329369 (i.e, number of permutations of ${1,2,...,m}$ with excedance set constructed by taking $m-i$ ($0 < i < m$) if $b(i-1) = 1$ where $b(k)b(k-1)\cdots b(1)b(0)$ ($0 \leqslant k ...
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2
votes
0
answers
70
views
Property of a family of simple polynomials related to the A329369
Let $a(n)$ be A329369 (i.e., number of permutations of $\{1,2,\dotsc,m\}$ with excedance set constructed by taking $m-i$ ($0 < i < m$) if $b(i-1) = 1$ where $b(k)b(k-1)\cdots b(1)b(0)$ ($0 \...
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2
votes
0
answers
103
views
$R$-recursion for the A235129
Let $a(n)$ be A235129 i.e. an integer sequence such that its exponential generating function $A(x)$ satisfies
$$
A'(x) = 1 + A(x)\exp(A(x))
$$
The sequence begins with
$$
1, 1, 3, 12, 64, 424, 3358, ...
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2
votes
0
answers
70
views
Elementary recursion for the A258173
Let $a(n)$ be A258173 i.e. sum over all Dyck paths of semilength $n$ of products over all peaks $p$ of $y_p$, where $y_p$ is the $y$-coordinate of peak $p$.
A Dyck path of semilength $n$ is a $(x,y)$-...
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2
votes
0
answers
90
views
Unexpected recursion for the A193231 (blue code of $n$)
Let $a(n)$ be A193231, blue code of $n$ i.e. self-inverse permutation of non-negative integers such that $a(n)<2^k$ iff $n<2^k$ and
$$
a(n\operatorname{XOR}k) = a(n) \operatorname{XOR} a(k)
$$
...
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2
votes
0
answers
72
views
Recursion for the number of partitions of $m^n-1$ into powers of $m$
Let $a(n,m)$ be the number of partitions of $m^n-1$ into powers of $m$. In other words,
$$a(n,m)=[z^{m^n-1}] \prod\limits_{k\geqslant 0} \frac{1}{1-z^{m^k}}$$
Let
$$
R(n,m,q)=\sum\limits_{j=0}^{m(q+1)-...
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2
votes
0
answers
126
views
Recurrence for A004208
Let $a(n)$ be A004208. Here
$$a(n)=n\prod\limits_{j=1}^{n}(2j-1)-\sum\limits_{i=1}^{n-1}a(i)\prod\limits_{j=1}^{n-i}(2j-1)$$
I conjecture that
$$a(n)=R(n-1,0)$$
where
$$R(n,q)=2(q+2)R(n-1,q+1)+\sum\...
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2
votes
0
answers
105
views
Sequences that sum up to the many sequences in the OEIS
Let
$$a(n,m,k)=\frac{1}{n}\sum\limits_{j=0}^{n}[n+kj\geqslant 0]\binom{n}{j}\binom{n+kj}{j-1}(m-1)^{j-1}$$
Here square brackets denote Iverson brackets.
There are many sequences in the OEIS that are ...
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2
votes
0
answers
116
views
Closed form for coefficients related to excedance set of permutation
Working on suitable closed form for A329369, I discovered very useful coefficients, which have the following recurrence relation:
$$T(0,1)=T(0,2)=1$$
$$T(n,1)=1, n>0$$
$$T(0,k)=0, k>2$$
$$T(2n+1,...
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2
votes
0
answers
61
views
Odious twin locations related to the sequence based on $d(n) = n-d(d(n-1))-d(d(n-2))$
Let $\operatorname{wt}(n)$ be A000120, i.e. the number of $1$'s in binary expansion of $n$ (or the binary weight of $n$).
Let $a(n)$ be the sequence of numbers $k$ such that $\operatorname{wt}(k)\...
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2
votes
0
answers
254
views
Conjecture on a sieve of Flavius Josephus
Flavius Josephus's sieve: Start with the natural numbers; at the $k$-th sieving step, remove every $(k+1)$-st term of the sequence remaining after the $(k-1)$-st sieving step; iterate.
Some examples:
...
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2
votes
0
answers
94
views
Subsequence of Laguerre polynomials
Let $m\geqslant1$ be a fixed integer.
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 $...
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1
vote
0
answers
32
views
On a A347205 and related row polynomials
Let $a(n)$ be A347205. Here
$$
a(2^m(2k+1)) = \sum\limits_{j=0}^{m}a(2^j k), \\
a(0) = 1.
$$
Let $\nu_2(n)$ be A007814 (i.e., number of trailing zeros in the binary expansion of $n$). Here
$$
\nu_2(2n+...
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1
vote
0
answers
63
views
On a A162326 and vector $\nu$ of length $n$
Let $a(n)$ be A162326. Here
$$
a(n) = \frac{1}{n}(2(5n-7)a(n-1) - 9(n-2)a(n-2)), \\
a(0) = a(1) = 1.
$$
Also ordinary generating function is
$$
\frac{5 - \sqrt{\frac{1-9x}{1-x}}}{4}.
$$
Let $b(n)$ be $...
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1
vote
0
answers
168
views
Integer coefficients and integrals
Let $a(n,p,q)$ be the family of integer sequences such that exponential generating functions for it satisfy
$$
A_1(x)=\exp\left(x + p\int\int (A_1(x))^q \, dx \, dx\right).
$$
Let $b(n,p,q)$ be the ...
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1
vote
0
answers
90
views
Closed form for the A357990 using A329369 and generalised A373183
Let
$$
\ell(n) = \left\lfloor\log_2 n\right\rfloor, \\
\ell(0) = -1
$$
Let
$$
f(n) = \ell(n) - \ell(n-2^{\ell(n)}) - 1
$$
Here $f(n)$ is A290255.
Let $A(n,k)$ be a square array such that
$$
A(n,k)...
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1
vote
0
answers
58
views
Simple recursion for the A329369 using Stirling numbers of both kinds
Let $s(n,k)$ be a (signed) Stirling number of the first kind.
Let $n \brace k$ be a Stirling number of the second kind.
Let $a(n)$ be A329369 (i.e, number of permutations of ${1,2,...,m}$ with ...
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1
vote
0
answers
132
views
Sequence that sums up to A000153
Let $a(n)$ be A329369 (i.e, number of permutations of ${1,2,...,m}$ with excedance set constructed by taking $m-i$ ($0 < i < m$) if $b(i-1) = 1$ where $b(k)b(k-1)\cdots b(1)b(0)$ ($0 \leqslant k ...
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1
vote
0
answers
105
views
Simpler recursion for the A358612
Let $T(n,k)$ be an integer coefficients (A358612) such that
$$
T(2n+1, k) = kT(n, k) + T(n, k-1), \\
T(2n, k) = kT(n, k) + T(n, k-1) - \frac{T(2n, k-1) + T(n, k-1)}{k-1}, \\
T(n, 1) = T(0, 2) = 1
$$
...
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1
vote
0
answers
73
views
Alternating sum of integer coefficients of the triangles related to Eulerian numbers and binomial transforms
Let $W(n, k, m)$ be an integer coefficients defined for $n > 0, 1 \leqslant k \leqslant n, m > 0$ with $W(n,k,m)=0$ for $n \leqslant 0$ or $k \leqslant 0$ such that
$$
W(n, k, m) = (k+m-1)W(n-1,...
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1
vote
0
answers
68
views
On a numbers $k$ with specific $2$-adic valuation
Let $a(n)$ be A002326 (i.e., multiplicative order of $2 \operatorname{mod} 2n+1$).
Let $b(n)$ be A179382 (i.e., the smallest period of pseudo-arithmetic progression with initial term $1$ and ...
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1
vote
0
answers
49
views
$R$-recursion for the A036765
Let $a(n)$ be A036765 i.e. number of ordered rooted trees with $n$ non-root nodes and all outdegrees $\leqslant 3$. Here
$$
a(n) = \frac{1}{n+1}\sum\limits_{j=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\...
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1
vote
0
answers
69
views
Simplification of computing $f(n,z)$
Let
$$
s(n,z)=\sum\limits_{j=0}^{n}L(n,j,z)
$$
where
$$
L(n,j,z)=\sum\limits_{p=0}^{n-j-1}f(p,z)L(n-j-1,p,z), \\
L(n,n,z)=1
$$
Now let $s(n,z)$ be an arbitrary function such that $s(0, z)=1$. It means ...
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1
vote
0
answers
103
views
Mysterious recursion for the A005225
Let $a(n)$ be A005225 i.e. number of permutations of length $n$ with equal cycles. Here
$$
a(n)=n!\sum\limits_{d|n}\frac{1}{d!(\frac{n}{d})^d}
$$
Let
$$
R(n,q,z)=(q+1)R(n-1,q+1,z)+\sum\limits_{j=0}^{q}...
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1
vote
0
answers
89
views
Suitable recursion for the A234289
Let $a(n)$ be A234289 i.e. integer sequence with exponential generating function
$$
A(x)=1+A(x)^2\int \frac{1}{A(x)}\,dx
$$
The sequence begins with
$$
1, 1, 3, 17, 147, 1729, 25827, 468593, 10012083, ...
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1
vote
0
answers
95
views
Pretty simple recursion for the A290383
Let $a(n)$ be A290383 i.e. number of set partitions of $[n]$ such that the smallest element of each block is odd. Here
$$
a(n)=b(n,0,0)
$$
where
$$
b(n,m,t)=\sum\limits_{j=1}^{m-t+1}b(n-1,\max(m,j),1-...
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1
vote
0
answers
80
views
Recursion for the A006014 using difference of binomial coefficients
Let $a(n)$ be A006014 i.e.
$$
a(n)=na(n-1)+\sum\limits_{j=1}^{n-2}a(j)a(n-j-1), \\
a(1)=1
$$
Also generating function $A(x)$ satisfies
$$
A(x) = x(1 + A(x) + A(x)^2 + xA'(x))
$$
Let
$$
R(n,q)=\sum\...
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1
vote
0
answers
111
views
Recursion for the Bessel polynomial $y_n(x)$
Let $a(n)$ be A001515 i.e. the Bessel polynomial $y_n(x)$ evaluated at $x=1$. Here
$$
a(n) = (2n-1)a(n-1) + a(n-2), \\
a(0) = 1, a(1) = 2
$$
The closed form is
$$
a(n)=\sum\limits_{k=0}^{n}\binom{n+k}{...
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1
vote
0
answers
94
views
Combinatorial interpretation for the more general case of $R(n,0)$
Let $f(n), g(n,m), h(n)$ be an arbitrary functions which equal to the non-negative integers.
Let
$$
R(n,q) = \sum\limits_{j=0}^{f(q)}g(q,j)R(n-1,j),\\
R(0,q) = h(q)
$$
In the comment to the one of ...
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1
vote
0
answers
93
views
Application of the series reversion
Let $f(n)$ be an arbitrary function such that $f(n)\in\mathbb{Z}$.
Let $a(n)$ be an arbitrary integer sequence such that $a(0)=1$.
Let $b(n)$ be an integer sequence such that
$$b(2^m(2n+1))=\sum\...
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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
$$\ell(n)=\left\lfloor\log_2 n\right\rfloor$$
$$f(n)=n-2^{\ell(n)}$$...
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1
vote
0
answers
56
views
Recurrence for the number of permutations with a given excedance set
Let
$$\ell(n)=\left\lfloor\log_2 n\right\rfloor$$
$$f(n)=n-2^{\ell(n)}$$
$$T(n,k)=\left\lfloor\frac{n}{2^k}\right\rfloor\operatorname{mod}2$$
$$\operatorname{wt}(2n+1)=\operatorname{wt}(n)+1, \...
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1
vote
0
answers
134
views
Recurrence for the A284005
Let
$$\ell(n)=\left\lfloor\log_2 n\right\rfloor$$
$$f(n)=n-2^{\ell(n)}$$
$$T(n,k)=\left\lfloor\frac{n}{2^k}\right\rfloor\operatorname{mod}2$$
$$\operatorname{wt}(2n+1)=\operatorname{wt}(n)+1, \...
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1
vote
0
answers
85
views
Permutation to get Stolarsky representation from lazy Fibonacci (dual Zeckendorf) representation
Let $a_1(n)$ be A200714, i.e., Stolarsky representation interpreted as binary to decimal integers. The sequence begins with
$$0, 1, 3, 2, 7, 5, 6, 15, 4, 11, 13, 14, 31, 10, 9, 23, 12, 27, 29$$
Let $...
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