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Results tagged with oeis
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user 231922
The acronym OEIS stands for the On-Line Encyclopedia of Integer Sequences, a well-known database of sequences of integers. It can be used for questions where this database is (or might be) relevant, mainly questions about particular sequences of integers. This tag is typically used in combination with other tags to make the scope of the question more precise; common examples of such tags include the top-level tags co.combinatorics and nt.number-theory.
0
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1
answer
172
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Binary recurrence from general recurrence
We have general recurrence for A243499 (which is product of parts of integer partitions as enumerated in the table A125106)
$$a(n)=(1+b(n))a(t(n)), a(0)=1$$
where $b(n)$ is A023416 (which is number of …
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1
vote
0
answers
191
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 co …
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1
vote
0
answers
91
views
Recurrence for the viabin numbers of the self-conjugate integer partitions
Let $a(n)$ be A290254, the viabin numbers of the self-conjugate integer partitions, also defined as $\left\lbrace 0 \right\rbrace$ union fixed points of A059894, self-inverse permutation defined as fo …
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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 $c(n)$ …
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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" bit. …
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0
votes
0
answers
106
views
Formula for individual term of the Proth numbers
The sequence begins with
$$
1, 1, 2, 3, 2, 3, 4, 5, 6, 7, 4, 5, 6, 7, 8
$$
There are no formulas for $b_1(n)$ and $b_2(n)$ in the OEIS, so after a little inspection I conjecture that
$$
c(n) = 2^{\ell( …
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5
votes
0
answers
118
views
Formula and smallest solution for the A260711
The sequence begins with
$$
1, 1, 2, 3, 2, 3, 4, 5, 6, 7, 4, 5, 6, 7, 8
$$
There are no formulas for $b_1(n)$ and $b_2(n)$ in the OEIS, so after a little inspection I conjecture that
$$
c(n) = 2^{\ell( …
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2
votes
0
answers
160
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 wit …
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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
$ …
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2
votes
1
answer
129
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
$$
a(n) = \frac{1}{2(n+1)}\sum\limits_{k=0}^{n}(k+1)((-1)^{\left\lflo …
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1
vote
0
answers
31
views
On a A347205 and related row polynomials
Let $a(n)$ be A347205. Here
$$
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(2 …
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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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1
vote
0
answers
120
views
Simple algorithm for A107670
Let $T(n, k)$ be A107670 (i.e., matrix square of triangle A107667). Here we define the triangular matrix $P$ by $P(n, k) = \frac{(n+1)^{2(n-k)}}{(n-k)!}$ for $0 \leqslant k \leqslant n$ and the diago …
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4
votes
1
answer
125
views
Intersecting algorithm for A065601
Let $a(n)$ be A065601 (i.e., number of Dyck paths of length $2n$ with exactly $1$ hill). Here
$$
a(n) = \frac{1}{2(n+1)}((3n-2)a(n-1) + 2(9n-19)a(n-2) + 4(2n-3)a(n-3)), \\
a(0) = a(2) = 0, a(1) = 1.
…
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4
votes
1
answer
106
views
On a number of compositions of $n$ into positive triangular numbers
Let $a(n)$ be A023361 (i.e., number of compositions of $n$ into positive triangular numbers). Here
$$
a(n) = \sum\limits_{i \geqslant 1, \frac{i(i+1)}{2}\leqslant n} a(n-\frac{i(i+1)}{2}), \\
a(0) = …
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