All Questions
15 questions
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 ...
- 4,938
7
votes
0
answers
226
views
Is there a connection between the sequence of a finite number of Stieltjes constants and the integer partitions number?
Lehmer 1988 and Keiper 1992 made major progress on evaluating the series:
$$\sigma_r = \sum_{k=1}^{\infty} \left( \frac{1}{\rho_k^r} + \frac{1}{(1-\rho_k)^r}\right) \quad r \in \mathbb{N}$$
where $\...
- 4,169
5
votes
1
answer
374
views
Closed-form for the number of partitions of $n$ avoiding the partition $(4,3,1)$
Let $a(n)$ be A309099 i.e. the number of partitions of $n$ avoiding the partition $(4,3,1)$.
We say a partition $\alpha$ contains $\mu$ provided that one can delete rows and columns from (the Ferrers ...
- 4,938
5
votes
0
answers
765
views
Computability of OEIS A034891 ...partitions of n into prime parts (1 included)
On the seqfan mailing list RGWv gave short algorithm for computing A000041 number of partitions of n the partition numbers:
f[1, 1] = 1; f[n_, k_] := f[n, k] = If[n < 0, 0, If[k > n, 0, If[k == ...
- 25.4k
4
votes
1
answer
206
views
Partition numbers as the specific sums of the A161511
Let $p(n)$ be A000041 i.e. number of partitions of $n$ (the partition numbers).
Let
$$
\ell(n)=\left\lfloor\log_2 n\right\rfloor
$$
Let $a(n)$ be A161511 i.e. number of $1\cdots0$ pairs in the ...
- 4,938
4
votes
1
answer
308
views
3 divides coefficents of this $q$-series
Denote $\phi(q):=\prod_{j\geq1}(1-q^j)$ and let $\xi=e^{\frac{2\pi i}3}$ be a cube root of unity.
Define the sequence $u(n)$ by
$$\prod_{n\geq1}\prod_{s=1}^2(1-q^n\xi^{ns})(1-q^{2n}\xi^{ns})
=\sum_{n\...
- 43.2k
4
votes
1
answer
206
views
Prime numbers and number of partitions of $n$ into distinct parts with boundary size $2$
Let $a(n)$ be A227559, i.e., number of partitions of $n$ into distinct parts with boundary size $2$. Be careful here: offset is $3$.
I conjecture that $a(4n+2)=2n+1$ for $n>0$ if and only if $2n+1$ ...
- 4,938
3
votes
1
answer
92
views
Partition of $(2^{n+1}+1)2^{2^{n-1}+n-1}-1$ into parts with binary weight equals $2^{n-1}+n$
Let $\operatorname{wt}(n)$ be A000120, i.e., number of $1$'s in binary expansion of $n$ (or the binary weight of $n$).
Let $a(n,m)$ be the sequence of numbers $k$ such that $\operatorname{wt}(k)=m$. ...
- 4,938
3
votes
0
answers
120
views
Sequence which is related to the binary expansion of $n$ and partition numbers
Let $p(n)$ be A000041 i.e. the number of partitions of $n$ (the partition numbers).
Let
$$
\ell(n)=\left\lfloor\log_2 n\right\rfloor
$$
Let $\operatorname{wt}(n)$ be A000120 i.e. number of $1$'s in ...
- 4,938
2
votes
1
answer
140
views
Number of branches between two layers of the Young's lattice
In the Young's lattice, the number of branches that connect the $N$'th layer to the $N+1$'th layer has the sequence:
$$
1,2, 4, 7, 12, 19, 30, 45, 67, 97, 139, \cdots
$$
Looking this up on OEIS, leads ...
- 153
2
votes
1
answer
164
views
Yet another question about unrestricted partitions
I posed a question called "A Product Related to Unrestricted Partitions". As it stands it is too hard. Here's another variation which is easier to search for and hopefully might shed some light on ...
- 2,143
2
votes
0
answers
72
views
Recursion for the number of partitions of $m^n-1$ into powers of $m$
Let $a(n,m)$ be the number of partitions of $m^n-1$ into powers of $m$. In other words,
$$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)-...
- 4,938
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-...
- 4,938
1
vote
0
answers
100
views
Conjecture on numbers $k$ having only one partition into parts with same binary weight as a binary weight of $k$
Let $\operatorname{tr}(n)$ be A007814, number of trailing zeros in the binary representation of $n$.
Also, let $\operatorname{ntr}(n)$ be A086784, number of non-trailing zeros in the binary ...
- 4,938
0
votes
0
answers
186
views
Are the numbers $\sum_{n=1}^\infty\frac1{p(n)}$ and $\sum_{n=1}^\infty\frac1{q(n)}$ transcendental?
For each positive integer $n$, let $p(n)$ be the number of partitions of $n$ (i.e., the number of ways to write $n$ as a sum of positive integers), and let $q(n)$ be the number of strict partitions of ...
- 15.6k