All Questions
10 questions
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
$$
a(n) = \frac{1}{2(n+1)}\sum\limits_{k=0}^{n}(k+1)((-1)^{\left\...
0
votes
0
answers
190
views
On a A057985 without recursion
Let $a(n)$ be A057985 (i.e., start with $0$ and repeatedly substitute: $0\to01, 1\to12, 2\to0$).
Let $\operatorname{wt}(n)$ be A000120 (i.e., number of ones in the binary expansion of $n$). Here
$$
\...
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 ...
35
votes
8
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 ...
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
$$
8, 16, 32, 48, 64, ...
0
votes
0
answers
107
views
Formula for individual term of the Proth numbers
Let $a(n)$ be A080075 i.e. Proth numbers: of the form $k2^m + 1$ for $k$ odd, $m \geqslant 1$ and $2^m > k$.
The sequence begins with
$$
3, 5, 9, 13, 17, 25, 33, 41, 49, 57, 65, 81, 97, 113, 129
$$...
2
votes
0
answers
72
views
Possible subsequence of the A110978
Let $a(n)$ be A110978 i.e. odd integers that are nonprime, such that there exist two factors of each number that when multiplied together in binary base, do not ever require the use of a "carry&...
1
vote
0
answers
100
views
Subsequence such that $c(a(n))=2^n$
Let $a(n)$ be A060831, i.e., $\sum\limits_{k=1}^{n}\operatorname{number of odd divisors of} k$.
Let
$$\ell(n)=\left\lfloor\log_2 n\right\rfloor$$
Let
$$b(n,k)=2b(n,k-1)-2^{k-1}, b(n,0)=n$$
Let $c(n)$ ...
1
vote
0
answers
194
views
Closed form for partial sums of A103318
Let $a(n)$ be A103318, number of solutions $i$ in range $[0,n-1]$ to $i \equiv 0 \pmod {2^{n-i}}$: the sequence begins with
$$1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 3, 1, 2, 2, 2, 1, 2, 2, 3, 2$$
Also let's ...
5
votes
1
answer
345
views
Why does this "factorial sequence" appear in the OEIS?
For a reciprocal of a polynomial, $f = \frac{1}{p}$, we (presumably) may construct a sequence $(c_n)_{n=0}^\infty$ such that for all $N\ge 0$
$$f(k)k! = \sum_{n=0}^{N-1} c_n(k-n)! + O((k-N)!). $$
I ...