All Questions
31 questions
2
votes
0
answers
119
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
$$
(x)_n = x(x-1)(x-2)\...
- 4,938
2
votes
0
answers
182
views
Algorithm for $\frac{1}{1-x} = \sum\limits_{n=0}^{\infty}a(n)x^n\prod\limits_{k=1}^{n}\frac{1-kx}{1+kx}$
Let $a(n)$ be A208832. Here
$$
\frac{1}{1-x} = \sum\limits_{n=0}^{\infty}a(n)x^n\prod\limits_{k=1}^{n}\frac{1-kx}{1+kx}.
$$
Start with vector $\nu$ of fixed length $m$ with elements $\nu_i = 1$ (that ...
- 4,938
2
votes
0
answers
67
views
$R$-recursion for A006351
Let $a(n)$ be A006351 (i.e., number of series-parallel networks with n labeled edges. Also called yoke-chains by Cayley and MacMahon). Here exponential generating function is $A(x)$ such that $B(x) = ...
- 4,938
2
votes
0
answers
59
views
$R$-recursion for A338193
Let $a(n)$ be A338193. Here generating function is $A(x)$ such that
$$
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 &...
- 4,938
1
vote
0
answers
161
views
Efficient algorithm for A217061
Let $a(n)$ be A217061. Here
$$
a(n) = \sum\limits_{m=1}^{n}\frac{1}{(m-1)!}\sum\limits_{k=0}^{n-m}(n+k-1)!\sum\limits_{j=0}^{k}\frac{1}{(k-j)!}\sum\limits_{\ell=0}^{j}\frac{2^{\ell-j}(-1)^{\ell+j}s(n-...
- 4,938
1
vote
1
answer
92
views
Equivalence of sequences related to A033264
Let $a(n)$ be A033264 (i.e., number of blocks of $\{1,0\}$ in the binary expansion of $n$). Here
$$
a(4n) = a(4n+1) = a(2n), \\
a(4n+2) = a(n)+1, \\
a(4n+3) = a(n), \\
a(0) = 0.
$$
Let
$$
\ell(n) = \...
- 4,938
1
vote
0
answers
89
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 ...
- 4,938
0
votes
0
answers
65
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 $(x,y)$-...
- 4,938
0
votes
0
answers
60
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 $\nu$ of ...
- 4,938
1
vote
1
answer
178
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
$$
a(n) = \sum\limits_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor} \binom{n-k-1}{...
- 4,938
2
votes
0
answers
51
views
Recursion for A129179 similar to recursion for Pascal's triangle
Let $T(n,k)$ be A129179 (i.e., triangle read by rows: $T(n, k)$ is the number of Schroeder paths of semilength $n$ such that the area between the $x$-axis and the path is $k$ ($n \geqslant 0, 0 \...
- 4,938
3
votes
1
answer
178
views
Algorithm for the sum with binomial coefficients and Bell numbers
Let $a(n)$ be A000110 (i.e., Bell or exponential numbers: number of ways to partition a set of $n$ labeled elements).
Let $b(n)$ be A355247 (i.e., expansion of exponential generating function $\exp(2(\...
- 4,938
4
votes
1
answer
112
views
On a number of compositions of $n$ into positive triangular numbers
Let $a(n)$ be A023361 (i.e., number of compositions of $n$ into positive triangular numbers). Here
$$
a(n) = \sum\limits_{i \geqslant 1, \frac{i(i+1)}{2}\leqslant n} a(n-\frac{i(i+1)}{2}), \\
a(0) = 1....
- 4,938
1
vote
0
answers
121
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 ...
- 4,938
4
votes
1
answer
130
views
Intersecting algorithm for A065601
Let $a(n)$ be A065601 (i.e., number of Dyck paths of length $2n$ with exactly $1$ hill). Here
$$
a(n) = \frac{1}{2(n+1)}((3n-2)a(n-1) + 2(9n-19)a(n-2) + 4(2n-3)a(n-3)), \\
a(0) = a(2) = 0, a(1) = 1.
$$...
- 4,938
3
votes
0
answers
165
views
Elegant algorithm for A140717
Let $T(n, k)$ be A140717 (i.e., triangle read by rows: $T(n,k)$ is the number of Dyck paths $d$ of semilength $n$ such that sum of peakheights of $d$ - number of peaks of $d$ equals $k$ ($n \geqslant ...
- 4,938
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
$$
\nu_2(2n+...
- 4,938
2
votes
1
answer
131
views
Sequence that sums up to A224071
Let $a(n)$ be A224071 (i.e., number of Schroeder paths of semilength $n$ in which there are no $(2,0)$-steps at level $1$). Here
$$
a(n) = \frac{1}{2(n+1)}\sum\limits_{k=0}^{n}(k+1)((-1)^{\left\...
- 4,938
0
votes
0
answers
190
views
On a A057985 without recursion
Let $a(n)$ be A057985 (i.e., start with $0$ and repeatedly substitute: $0\to01, 1\to12, 2\to0$).
Let $\operatorname{wt}(n)$ be A000120 (i.e., number of ones in the binary expansion of $n$). Here
$$
\...
- 4,938
2
votes
0
answers
163
views
Interesting conjecture by Sequence Machine
Let $a(n)$ be A344960 (i.e., position of binary complement of $n$-th word in A341258). By definition, in order to calculate $a(n)$, we need to know A341258. Below we will correspond this sequence with ...
- 4,938
5
votes
0
answers
133
views
Formula and smallest solution for the A260711
Let $a(n)$ be A260711 without initial $0$ (i.e., numbers of the form $x^2 - y^2$ with $x > y$ where $x$ and $y$ are odd, $x + y$ is a power of $2$).
The sequence begins with
$$
8, 16, 32, 48, 64, ...
- 4,938
0
votes
0
answers
107
views
Formula for individual term of the Proth numbers
Let $a(n)$ be A080075 i.e. Proth numbers: of the form $k2^m + 1$ for $k$ odd, $m \geqslant 1$ and $2^m > k$.
The sequence begins with
$$
3, 5, 9, 13, 17, 25, 33, 41, 49, 57, 65, 81, 97, 113, 129
$$...
- 4,938
2
votes
0
answers
72
views
Possible subsequence of the A110978
Let $a(n)$ be A110978 i.e. odd integers that are nonprime, such that there exist two factors of each number that when multiplied together in binary base, do not ever require the use of a "carry&...
- 4,938
1
vote
0
answers
100
views
Subsequence such that $c(a(n))=2^n$
Let $a(n)$ be A060831, i.e., $\sum\limits_{k=1}^{n}\operatorname{number of odd divisors of} k$.
Let
$$\ell(n)=\left\lfloor\log_2 n\right\rfloor$$
Let
$$b(n,k)=2b(n,k-1)-2^{k-1}, b(n,0)=n$$
Let $c(n)$ ...
- 4,938
1
vote
0
answers
193
views
Closed form for partial sums of A103318
Let $a(n)$ be A103318, number of solutions $i$ in range $[0,n-1]$ to $i \equiv 0 \pmod {2^{n-i}}$: the sequence begins with
$$1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 3, 1, 2, 2, 2, 1, 2, 2, 3, 2$$
Also let's ...
- 4,938
13
votes
2
answers
674
views
A reformulation of Erdős conjecture on arithmetic progressions
Erdős conjecture on arithmetic progressions states that if $S$ is a set of positive integers such that $c(S):=\sum_{n \in S} \frac{1}{n} = \infty$ (large set), then $ \forall \ell \ge 3$ the set $S$ ...
9
votes
0
answers
225
views
On the first sequence without collinear triple
Let $u_n$ be the sequence lexicographically first among the sequences of nonnegative integers with graphs without collinear three points (as for $a_n=n^2$ or $b_n=2^n$). It is a variation of that one.
...
57
votes
0
answers
3k
views
On the first sequence without triple in arithmetic progression
In this Numberphile video (from 3:36 to 7:41), Neil Sloane explains an amazing sequence:
It is the lexicographically first among the sequences of positive integers without triple in arithmetic ...
33
votes
7
answers
3k
views
Examples of integer sequences coincidences
For the time being, the OEIS website contains almost $300000$ sequences. Each of these sequences is the mark of a specific mathematical concept. Sometimes two (or more) distinct concepts have the ...
42
votes
2
answers
2k
views
Numbers that are generic w.r.t. exponentiation
This is a follow-up to my old question Number of distinct values taken by $x\hat{\phantom{\hat{}}}x\hat{\phantom{\hat{}}}\dots\hat{\phantom{\hat{}}}x$ with parentheses inserted in all possible ways.
...
- 6,819
19
votes
1
answer
1k
views
Number of distinct values taken by x^x^...^x with parentheses inserted in all possible ways
For what positive x's the number of distinct values taken by x^x^...^x with parentheses inserted in all possible ways is not represented by the sequence A000081? Is it exactly the set of positive ...