All Questions
14 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
$$
a(n) = \frac{1}{2(n+1)}\sum\limits_{k=0}^{n}(k+1)((-1)^{\left\...
- 4,948
0
votes
0
answers
190
views
On a A057985 without recursion
Let $a(n)$ be A057985 (i.e., start with $0$ and repeatedly substitute: $0\to01, 1\to12, 2\to0$).
Let $\operatorname{wt}(n)$ be A000120 (i.e., number of ones in the binary expansion of $n$). Here
$$
\...
- 4,948
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 ...
- 4,948
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
$$
8, 16, 32, 48, 64, ...
- 4,948
0
votes
0
answers
107
views
Formula for individual term of the Proth numbers
Let $a(n)$ be A080075 i.e. Proth numbers: of the form $k2^m + 1$ for $k$ odd, $m \geqslant 1$ and $2^m > k$.
The sequence begins with
$$
3, 5, 9, 13, 17, 25, 33, 41, 49, 57, 65, 81, 97, 113, 129
$$...
- 4,948
2
votes
0
answers
72
views
Possible subsequence of the A110978
Let $a(n)$ be A110978 i.e. odd integers that are nonprime, such that there exist two factors of each number that when multiplied together in binary base, do not ever require the use of a "carry&...
- 4,948
1
vote
0
answers
100
views
Subsequence such that $c(a(n))=2^n$
Let $a(n)$ be A060831, i.e., $\sum\limits_{k=1}^{n}\operatorname{number of odd divisors of} k$.
Let
$$\ell(n)=\left\lfloor\log_2 n\right\rfloor$$
Let
$$b(n,k)=2b(n,k-1)-2^{k-1}, b(n,0)=n$$
Let $c(n)$ ...
- 4,948
0
votes
1
answer
104
views
Non-Wieferich primes with Euler quotient modulo $p$ two and alternating harmonic numbers
Let $b(n)$ denote the Euler quotient modulo $n$.
In OEIS we have A128465 Numbers k such that k divides the numerator of alternating Harmonic number H'((k+1)/2)
For $n>1$ we have $b(A128465(n))=2$.
...
- 25.4k
1
vote
0
answers
194
views
Closed form for partial sums of A103318
Let $a(n)$ be A103318, number of solutions $i$ in range $[0,n-1]$ to $i \equiv 0 \pmod {2^{n-i}}$: the sequence begins with
$$1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 3, 1, 2, 2, 2, 1, 2, 2, 3, 2$$
Also let's ...
- 4,948
2
votes
0
answers
327
views
Why can one compute the sum of divisors of $n$ without factoring $n$?
Question links to paper
which states:
$$
\sigma(n)= \frac{6}{n^2(n-1)}\sum_{k=1}^{n-1}(3n^2-10k^2)\sigma(k)\sigma(n-k) \qquad (1)
$$
where $\sigma(n)$ is the sum of divisors of $n$.
Another similar ...
- 25.4k
33
votes
2
answers
856
views
A sequence potentially consisting of only integers
I will first ask the question which can be stated very simply. Afterwards I will explain some motivation and give references to related sequences.
Consider the sequence defined by
$$b_n = \frac{(...
- 7,875
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 ...
19
votes
1
answer
1k
views
Is OEIS A007018 really a subsequence of squarefree numbers?
A comment in A007018 a(n) = a(n-1)^2 + a(n-1), a(0)=1 claims
Subsequence of squarefree numbers (A005117). - Reinhard Zumkeller, Nov 15 2004
Is that really so?
As far as I know, it is an open ...
- 25.4k
11
votes
1
answer
864
views
Up to $10^6$: $\sigma(8n+1) \mod 4 = OEIS A001935(n) \mod 4$ (Number of partitions with no even part repeated )
Up to $10^6$:
$\sigma(8n+1) \mod 4 = OEIS A001935(n) \mod 4$
A001935 Number of partitions with no even part repeated
Is this true in general?
It would mean relation between restricted partitions ...
- 25.4k