All Questions
8 questions
2
votes
1
answer
216
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
$$
f(n,m,i) = (-1)^{m-i+1}\sum\limits_{j=i}^{m+1}j^n s(j,i) {m+1 \brace ...
- 4,948
1
vote
0
answers
115
views
Representing A329369 using A358612
Let $a(n)$ be A329369 (i.e., number of permutations of $\{1,2,\dotsc,m\}$ with excedance set constructed by taking $m-i$ ($0 < i < m$) if $b(i-1) = 1$ where $b(k)b(k-1)\cdots b(1)b(0)$ ($0 \...
- 4,948
4
votes
1
answer
148
views
Closed form for the A110501 (unsigned Genocchi numbers (of first kind) of even index)
Let $a(n)$ be A110501 (i.e., unsigned Genocchi numbers (of first kind) of even index). Here
$$
a(n) = \sum\limits_{i=1}^{\left\lfloor\frac{n}{2}\right\rfloor}\binom{n}{2i}a(n-i)(-1)^{i-1}, \\
a(1) = 1
...
- 4,948
1
vote
1
answer
114
views
Coefficients of number of the same terms which are arising from iterations based on binary expansion of $n$
Let
$$\ell(n)=\left\lfloor\log_2 n\right\rfloor$$
Let
$$T(n,k)=\left\lfloor\frac{n}{2^k}\right\rfloor\operatorname{mod}2$$
Here $T(n,k)$ is the $(k+1)$-th bit from the right side in the binary ...
- 4,948
2
votes
0
answers
157
views
Closed form for the A347205
Let $q(n)$ be A007814, i.e., number of trailing zeros in the binary representation of $n$. Here
$$q(2n+1)=0, q(2n)=q(n)+1$$
Let $\operatorname{wt}(n)$ be A000120, i.e., number of $1$'s in binary ...
- 4,948
2
votes
0
answers
115
views
Closed form for the sum of the integer coefficients
Let $a(n)$ be A002720, i.e., number of partial permutations of an $n$-set; number of $n \times n$ binary matrices with at most one $1$ in each row and column.
$$a(n)=\sum\limits_{k=0}^{n} k!\binom{n}{...
- 4,948
0
votes
0
answers
94
views
Closed form for the number of steps required to get $n$ balls in the last box
Let $\operatorname{wt}(n)$ be A000120, i.e., number of $1$'s in binary expansion of $n$ (or the binary weight of $n$).
Then we have an integer sequence given by
$$a(n)=n(n+1)-\sum\limits_{k=0}^{n}\...
- 4,948
3
votes
0
answers
165
views
Closed form for $a(2^m(2^n-2^p-1))$
Let $q(n)$ be A007814, i.e., the number of trailing zeros in the binary representation of $n$. Here
$$q(2n+1)=0, q(2n)=q(n)+1$$
Let $a(n)$ be A329369. Here
$$a(2n+1)=a(n), a(2n)=a(n)+a(n-2^{q(n)})+a(...
- 4,948