All Questions
259 questions
0
votes
1
answer
129
views
Closed form for $\sum\limits_{k=0}^{n} [\operatorname{wt}(k) = m]$ where $\operatorname{wt}(n)$ is the binary weight of $n$
Let $\operatorname{wt}(n)$ be A000120 (i.e., number of $1$'s in binary expansion of $n$).
Let $a(n,m)$ be the family of integer sequences such that
$$
a(n,m) = \sum\limits_{k=0}^{n} [\operatorname{wt}(...
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8
votes
1
answer
671
views
Infinite series and sum of two squares
Consider the following infinite sequence $a(n)$ generated by
$$\sum_{n\geq0} a(n)q^n
=\frac{\sum_{k\geq0}F(2k+1)q^{\binom{k+1}2}}{\sum_{k\geq0} q^{\binom{k+1}2}}$$
where the $F(2k+1)$ are the odd ...
- 43.2k
0
votes
1
answer
169
views
Partial sums of binomial coefficients and related family of polynomials
Let $a(n)$ be A302117. Here
$$
a(n) = 4(n-1)a(n-1) - \frac{1}{3}\prod\limits_{k=0}^{n-1}(2k-3), \\
a(0) = 0.
$$
Let
$$
T(n,k) = \sum\limits_{i=0}^{k} \binom{n}{i}.
$$
Let $P_n(z)$ be the family of ...
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1
vote
0
answers
57
views
Step back step forward algorithm for A108442
Let $a(n)$ be A108442. Here generating function is $\frac{1}{1-zA(z)}$ where
$$
A(z) = 1 + z(A(z))^2 + z(A(z))^3.
$$
Also
$$
a(n) = \sum\limits_{k=1}^{n}\frac{k}{2n-k}\sum\limits_{i=0}^{n-k} \binom{2n-...
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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 ...
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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
$$
R(n, q) = \begin{cases}
1 &...
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1
vote
0
answers
161
views
Efficient algorithm for A217061
Let $a(n)$ be A217061. Here
$$
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-...
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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
$$
\ell(n) = \...
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1
vote
0
answers
89
views
Test for odd prime triples in a $2p-1$ progression
Let $a(n)$ be A057326 (i.e., first member of a prime triple in a $2p-1$ progression).
Let $b(n) = B$ after $n-1$ iterations where we start with $A=n, B=1$ and for $i$ from $1$ to $n-1$ simultaneously ...
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2
votes
0
answers
113
views
Numbers of positive terms in polynomials equal A069999
Let $a(n)$ be A069999 (i.e., number of possible dimensions for commutators of $n \times n$ matrices; it is independent of the field). OEIS states that no generating function is known.
Let $P(n,k)$ be ...
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1
vote
0
answers
82
views
Generating functions related to generating function of Catalan numbers
Let $C_n$ be A000108 (i.e., Catalan numbers). Here generating function is $C(x)$ such that
$$
C(x) = \frac{1-\sqrt{1-4x}}{2x}.
$$
Let $a(n)$ be an integer sequence with generating function $A(x)$ such ...
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0
votes
0
answers
65
views
Algorithm and equivalent recursion for A258173 (related to Dyck paths)
Let $a(n)$ be A258173 i.e. sum over all Dyck paths of semilength $n$ of products over all peaks $p$ of $y_p$, where $y_p$ is the $y$-coordinate of peak $p$.
A Dyck path of semilength $n$ is a $(x,y)$-...
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1
vote
1
answer
178
views
Algorithm for A127782
Let $a(n)$ be A127782 (i.e., an integer sequence with generating function $A(x)$ such that $A(x)=1+xA(x+x^2)$). Here
$$
a(n) = \sum\limits_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor} \binom{n-k-1}{...
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23
votes
4
answers
2k
views
Identity for an infinite product
Here is an experimental "result" exhibiting the difference of two (formal) infinite products that "almost factorizes".
QUESTION. Is this true?
$$\prod_{n\geq1}(1+x^{2n-1})^{24} - \...
- 43.2k
2
votes
0
answers
61
views
Algorithm for main diagonal of integer coefficients associated with Schroeder numbers
Let $T_q(n, k)$ be an integer table such that
$$T_q(n, k) = \begin{cases}
1 & \textrm{if } n = 0 \vee k = 0 \\
qT_q(n-1, n-1) + T_q(n, n-1) & \textrm{if } n = k > 0 \\
T_q(n, k-1) + T_q(n-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
$$
A(x) = \cfrac{1}{1-x-\cfrac{x^2}{1-x^3-\cfrac{x^4}{1-x^5-\cfrac{x^6}{1-x^7-\cfrac{x^8}{\ddots}}}}}...
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2
votes
1
answer
310
views
Generating function for A300483 (related to Chebyshev polynomial of first kind)
Let $a(n)$ be A300483. Here
$$
a(n) = 2\int\limits_{t \geqslant 0}T_n\left(\frac{t+1}{2}\right)\exp(-t)\,dt.
$$
where $T_n(x)$ is $n$-th Chebyshev polynomial of first kind.
Let $b(n)$ be an integer ...
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10
votes
1
answer
625
views
Generating function for A261041
Let $a(n)$ be A261041 (i.e., number of partitions of subsets of $\{1,2,\dotsc,n\}$, where consecutive integers are required to be in different parts).
Let $b(n)$ be an integer sequence with generating ...
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3
votes
1
answer
178
views
Algorithm for the sum with binomial coefficients and Bell numbers
Let $a(n)$ be A000110 (i.e., Bell or exponential numbers: number of ways to partition a set of $n$ labeled elements).
Let $b(n)$ be A355247 (i.e., expansion of exponential generating function $\exp(2(\...
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4
votes
1
answer
112
views
On a number of compositions of $n$ into positive triangular numbers
Let $a(n)$ be A023361 (i.e., number of compositions of $n$ into positive triangular numbers). Here
$$
a(n) = \sum\limits_{i \geqslant 1, \frac{i(i+1)}{2}\leqslant n} a(n-\frac{i(i+1)}{2}), \\
a(0) = 1....
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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
$$
a(n) = \frac{1}{2(n+1)}((3n-2)a(n-1) + 2(9n-19)a(n-2) + 4(2n-3)a(n-3)), \\
a(0) = a(2) = 0, a(1) = 1.
$$...
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6
votes
1
answer
282
views
Integer sequences with a periodic pattern
Let $A$ and $B$ be two different integers. Let $S$ be a finite integer sequence with exactly $n_A$ $A$s and $n_B$ $B$s. By repeating $S$ infinitely many times we obtain an infinite integer sequence $P$...
- 93
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 ...
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2
votes
0
answers
64
views
On a $\sum\limits_{n=0}^{\infty}c_n x^n=\sum\limits_{n=0}^{\infty}a(n)x^n\prod\limits_{k=1}^{n+1}(1-f(k)x^k)$ (slightly different question)
Please note that this question differs from one of the previous questions of mine.
Let $f(n)$ be an arbitrary function with integer values.
Let $c_n$ be an arbitrary integer sequence.
Let $a(n)$ be ...
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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
$$
\nu_2(2n+...
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5
votes
1
answer
168
views
On a generating function and vector $\nu$ of length $n$
Let $f(n)$ be an arbitrary function with integer values.
Let $a(n)$ be an integer sequence such that
$$
\frac{1}{1-x}=\sum\limits_{n=0}^{\infty}a(n)x^n\prod\limits_{k=1}^{n+1}(1-f(k)x)
$$
Start with ...
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9
votes
0
answers
258
views
On a continued fraction and vector $\nu$ of length $n$
Please note that this question has been completely reworked in order not to overload it with unnecessary and useless information.
Let $f(n)$ be an arbitrary function with integer values.
Let $a(n)$ ...
- 4,938
1
vote
0
answers
63
views
On a A162326 and vector $\nu$ of length $n$
Let $a(n)$ be A162326. Here
$$
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}.
$$
Let $b(n)$ be $...
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2
votes
2
answers
315
views
5 different ways to define the same family of integer sequences
Let ${n \brace k}$ be a Stirling number of the second kind.
Let $A_n(x)$ be an Eulerian polynomial. Here
$$
A_n(x) = \sum_{i=0}^{n}i!{n \brace i}(x-1)^{n-i}.
$$
Let $a_1(n,p,q)$ be the family of ...
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0
votes
1
answer
94
views
Do disjunctive sequences eventually get palindromic at some point?
I have a friend who is very interested in math and has been thinking about a problem involving disjunctive sequences. For his birthday, I would like to give him an answer to his question, either by ...
0
votes
0
answers
135
views
Integer coefficients and continued fractions
Let $a(n,p,q)$ be the family of integer sequences such that ordinary generating functions for it are $\frac{1}{G_1(0,x)}$ where $G_1(0,x)$ are continued fractions such that
$$
G_1(j,x)=1-\cfrac{(qj+1)...
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3
votes
1
answer
140
views
$R$-recursion for unsigned Genocchi numbers (of first kind) of even index
Let $G_n$ be A036968 (i.e., Genocchi numbers). Here
$$
\frac{2t}{1+e^t}=\sum\limits_{n=0}^{\infty}G_n\frac{t^n}{n!}.
$$
Also
$$
t\tan\left(\frac{t}{2}\right)=\sum\limits_{n=1}^{\infty}(-1)^n G_{2n}\...
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2
votes
1
answer
147
views
$R$-recursion for Fibonacci numbers using signed Catalan numbers
Let $F_n$ be A000045 (i.e., Fibonacci numbers). Here
$$
F_n = F_{n-1} + F_{n-2}, \\
F_0 = 0, F_1 = 1.
$$
Let $C_n$ be A000108 (i.e., Catalan numbers). Here
$$
C_n = \frac{1}{n+1}\binom{2n}{n}.
$$
Let
$...
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1
vote
1
answer
61
views
Simplest way to generate integer coefficients with row sums equal to the terms of an arbitrary given sequence
Let $f(n)$ be an arbitrary function.
Let $\operatorname{wt}(n)$ be A000120 (i.e., number of ones in the binary expansion of $n$). Here
$$
\operatorname{wt}(2n+1) = \operatorname{wt}(n) + 1, \\
\...
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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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0
votes
0
answers
55
views
Sequences that sum up to sums of integer coefficients
Let
$$
T(n,k,p,q,r,s) = (q(k-1)+1)T(n-1,k,p,q,r,s) + s(n+r(k-1)+p-2)T(n-1,k-1,p,q,r,s), \\
T(n,1,p,q,r,s) = 1, \\
T(n,0,p,q,r,s) = T(0,k,p,q,r,s) = 0
$$
Let
$$
\ell(n) = \left\lfloor\log_2 n\right\...
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1
vote
0
answers
90
views
Closed form for the A357990 using A329369 and generalised A373183
Let
$$
\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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6
votes
0
answers
752
views
For all $n\in \mathbb{N}$, How to find $\min\{ m+k\}$ such that $ \binom{m}{k}=n$?
I asked this question on MSE here.
Most numbers in pascal triangle appear only once (excluding the duplicates in the same row of the Pascal's triangle) but certain numbers appear multiple times. ...
- 541
1
vote
0
answers
175
views
Solution of recurrence relation with summation
I have the following recurrence relation:
$$b(n,k)=\sum _{\text{i}=0}^{2 n-1} \left(b(n-1,k-\text{i})+\frac{\text{i} (2 n-\text{i}) \binom{2 n-1}{\text{i}} \binom{(n-2)^2}{k-\text{i}}}{2 n-1} \right)$$...
- 47
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
$$
A(x) = \sum\limits_{k=0}^{\infty} \frac{a(k)}{...
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2
votes
1
answer
214
views
Simplification of the closed form for the A329369
Let $s(n,k)$ be a (signed) Stirling number of the first kind.
Let ${n \brace k}$ be a Stirling number of the second kind.
Let
$$
f(n,m,i) = (-1)^{m-i+1}\sum\limits_{j=i}^{m+1}j^n s(j,i) {m+1 \brace ...
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2
votes
1
answer
129
views
Recursion for the sum with 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
$$
f(n,m,i) = (-1)^{m-i+1}\sum\limits_{j=i}^{m+1}j^n s(j,i) {m+1 \brace j}...
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2
votes
0
answers
100
views
Another (unique) algorithm for the A329369
Let $a(n)$ be A329369 (i.e, number of permutations of ${1,2,...,m}$ with excedance set constructed by taking $m-i$ ($0 < i < m$) if $b(i-1) = 1$ where $b(k)b(k-1)\cdots b(1)b(0)$ ($0 \leqslant k ...
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1
vote
0
answers
85
views
Closed form for the family of polynomials
Let $s(n,k)$ be a (signed) Stirling number of the first kind.
Let $R(n,x)$ be the family of polynomials such that
$$
R(2n+1,x) = xR(n,x), \\
R(2n,x) = x(R(n,x+1) - R(n, x)), \\
R(0, x) = x
$$
Let $\...
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1
vote
0
answers
58
views
Simple recursion for the A329369 using Stirling numbers of both kinds
Let $s(n,k)$ be a (signed) Stirling number of the first kind.
Let $n \brace k$ be a Stirling number of the second kind.
Let $a(n)$ be A329369 (i.e, number of permutations of ${1,2,...,m}$ with ...
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1
vote
0
answers
132
views
Sequence that sums up to A000153
Let $a(n)$ be A329369 (i.e, number of permutations of ${1,2,...,m}$ with excedance set constructed by taking $m-i$ ($0 < i < m$) if $b(i-1) = 1$ where $b(k)b(k-1)\cdots b(1)b(0)$ ($0 \leqslant k ...
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3
votes
0
answers
88
views
Recursion for reversed rows of the A373183 using unsigned Stirling numbers of the first kind
Let $\left[{n \atop k}\right]$ be unsigned Stirling numbers of the first kind. Here
$$
\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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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
113
views
Representing A329369 using A358612
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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