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Results tagged with oeis
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user 231922
The acronym OEIS stands for the On-Line Encyclopedia of Integer Sequences, a well-known database of sequences of integers. It can be used for questions where this database is (or might be) relevant, mainly questions about particular sequences of integers. This tag is typically used in combination with other tags to make the scope of the question more precise; common examples of such tags include the top-level tags co.combinatorics and nt.number-theory.
2
votes
0
answers
117
views
Generalized identity with Stirling numbers of the second kind and falling factorials
It is known that Striling numbers of the second kind satisfy the relation
$$
\sum\limits_{k=0}^{n}{n \brace k}(x)_k = x^n.
$$
where $(x)_n$ is the falling factorials such that
$$
(x)_n = x(x-1)(x-2)\d …
- 4,938
2
votes
0
answers
136
views
Algorithm for $\frac{1}{1-x} = \sum\limits_{n=0}^{\infty}a(n)x^n\prod\limits_{k=1}^{n}\frac{...
Let $a(n)$ be A208832. Here
$$
\frac{1}{1-x} = \sum\limits_{n=0}^{\infty}a(n)x^n\prod\limits_{k=1}^{n}\frac{1-kx}{1+kx}.
$$
Start with vector $\nu$ of fixed length $m$ with elements $\nu_i = 1$ (that …
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2
votes
0
answers
50
views
$R$-recursion for A006351
Let $a(n)$ be A006351 (i.e., number of series-parallel networks with n labeled edges. Also called yoke-chains by Cayley and MacMahon). Here exponential generating function is $A(x)$ such that $B(x) = …
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2
votes
0
answers
47
views
$R$-recursion for A338193
Let $a(n)$ be A338193. Here generating function is $A(x)$ such that
$$
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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1
vote
0
answers
141
views
Efficient algorithm for A217061
Let $a(n)$ be A217061. Here
$$
a(n) = \sum\limits_{m=1}^{n}\frac{1}{(m-1)!}\sum\limits_{k=0}^{n-m}(n+k-1)!\sum\limits_{j=0}^{k}\frac{1}{(k-j)!}\sum\limits_{\ell=0}^{j}\frac{2^{\ell-j}(-1)^{\ell+j}s(n …
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1
vote
1
answer
64
views
Equivalence of sequences related to A033264
Let $a(n)$ be A033264 (i.e., number of blocks of $\{1,0\}$ in the binary expansion of $n$). Here
$$
a(4n) = a(4n+1) = a(2n), \\
a(4n+2) = a(n)+1, \\
a(4n+3) = a(n), \\
a(0) = 0.
$$
Let
$$
\ell(n) = \ …
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1
vote
0
answers
66
views
Test for odd prime triples in a $2p-1$ progression
Let $a(n)$ be A057326 (i.e., first member of a prime triple in a $2p-1$ progression).
Let $b(n) = B$ after $n-1$ iterations where we start with $A=n, B=1$ and for $i$ from $1$ to $n-1$ simultaneously …
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0
votes
0
answers
58
views
Algorithm and equivalent recursion for A258173 (related to Dyck paths)
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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0
votes
0
answers
43
views
Algorithm for $q$-Bell numbers
Let $T(n,k)$ be A126347 (i.e., triangle, read by rows, with row polynomials $B(n, q)$). Here
$$
B(n, q) = \sum\limits_{k=0}^{n-1}\binom{n-1}{k}B(k, q)q^k, \\
B(0, q) = 1.
$$
Start with vector $\nu$ o …
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1
vote
1
answer
167
views
Algorithm for A127782
Let $a(n)$ be A127782 (i.e., an integer sequence with generating function $A(x)$ such that $A(x)=1+xA(x+x^2)$). Here
$$
a(n) = \sum\limits_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor} \binom{n-k-1} …
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2
votes
0
answers
43
views
Recursion for A129179 similar to recursion for Pascal's triangle
Let $T(n,k)$ be A129179 (i.e., triangle read by rows: $T(n, k)$ is the number of Schroeder paths of semilength $n$ such that the area between the $x$-axis and the path is $k$ ($n \geqslant 0, 0 \leqs …
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3
votes
1
answer
166
views
Algorithm for the sum with binomial coefficients and Bell numbers
Let $a(n)$ be A000110 (i.e., Bell or exponential numbers: number of ways to partition a set of $n$ labeled elements).
Let $b(n)$ be A355247 (i.e., expansion of exponential generating function $\exp(2 …
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4
votes
1
answer
106
views
On a number of compositions of $n$ into positive triangular numbers
Let $a(n)$ be A023361 (i.e., number of compositions of $n$ into positive triangular numbers). Here
$$
a(n) = \sum\limits_{i \geqslant 1, \frac{i(i+1)}{2}\leqslant n} a(n-\frac{i(i+1)}{2}), \\
a(0) = …
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1
vote
0
answers
120
views
Simple algorithm for A107670
Let $T(n, k)$ be A107670 (i.e., matrix square of triangle A107667). Here we define the triangular matrix $P$ by $P(n, k) = \frac{(n+1)^{2(n-k)}}{(n-k)!}$ for $0 \leqslant k \leqslant n$ and the diago …
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4
votes
1
answer
125
views
Intersecting algorithm for A065601
Let $a(n)$ be A065601 (i.e., number of Dyck paths of length $2n$ with exactly $1$ hill). Here
$$
a(n) = \frac{1}{2(n+1)}((3n-2)a(n-1) + 2(9n-19)a(n-2) + 4(2n-3)a(n-3)), \\
a(0) = a(2) = 0, a(1) = 1.
…
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