All Questions
33 questions
2
votes
0
answers
76
views
upper and lower bounds on rowlands sequence
rowlands sequence is defined as follows
\begin{equation}
a_{n}=a_{n-1} + b_{n}
\end{equation}
where $b_{n} = gcd(a_{n-1}, n)$ for $n>h$
it originates from E. Rowlands 2008 paper "A Natural ...
1
vote
0
answers
168
views
Integer coefficients and integrals
Let $a(n,p,q)$ be the family of integer sequences such that exponential generating functions for it satisfy
$$
A_1(x)=\exp\left(x + p\int\int (A_1(x))^q \, dx \, dx\right).
$$
Let $b(n,p,q)$ be the ...
- 4,948
0
votes
0
answers
28
views
Short periods modulo primes of linear recurrences with polynomial coefficients
Let $f_i(x)$ be polynomials with integer coefficients.
Define the integer linear recurrence with polynomial coefficients:
$$
a(n)=f_1(n) a(n-1)+f_2(n)a(n-2)+\cdots +f_d(n) a(n-d)
$$
and the initial ...
- 25.4k
0
votes
0
answers
55
views
Sequences that sum up to sums of integer coefficients
Let
$$
T(n,k,p,q,r,s) = (q(k-1)+1)T(n-1,k,p,q,r,s) + s(n+r(k-1)+p-2)T(n-1,k-1,p,q,r,s), \\
T(n,1,p,q,r,s) = 1, \\
T(n,0,p,q,r,s) = T(0,k,p,q,r,s) = 0
$$
Let
$$
\ell(n) = \left\lfloor\log_2 n\right\...
- 4,948
1
vote
0
answers
133
views
Sequence that sums up to A000153
Let $a(n)$ be A329369 (i.e, number of permutations of ${1,2,...,m}$ with excedance set constructed by taking $m-i$ ($0 < i < m$) if $b(i-1) = 1$ where $b(k)b(k-1)\cdots b(1)b(0)$ ($0 \leqslant k ...
- 4,948
1
vote
0
answers
106
views
Simpler recursion for the A358612
Let $T(n,k)$ be an integer coefficients (A358612) such that
$$
T(2n+1, k) = kT(n, k) + T(n, k-1), \\
T(2n, k) = kT(n, k) + T(n, k-1) - \frac{T(2n, k-1) + T(n, k-1)}{k-1}, \\
T(n, 1) = T(0, 2) = 1
$$
...
- 4,948
6
votes
0
answers
245
views
Searching for a proof of the pattern and identification of integer coefficients for the A329369
Please see the update given below. Everything you need to know from the old version of the question are the functions $a(n), \ell(n), s(n), t(n), r(n)$.
Let $a(n)$ be A329369 (i.e, number of ...
- 4,948
6
votes
1
answer
367
views
On A057985 and A287066
Let $a(n)$ be A057985 (i.e., start with $0$ and repeatedly substitute: $0 \to 01$, $1 \to 12$, $2 \to 0$).
Let $b(n)$ be A287066 (i.e., start with $1$ and repeatedly substitute: $0 \to 01$, $1 \to 12$...
- 4,948
2
votes
1
answer
120
views
Recursion for the Chebyshev transform of $m^n$
Let
$$
R(n, q, m) = R(n-1, q+1, m) + \sum\limits_{j=0}^{q} (-1)^{q-j}R(n-1, j, m), \\
R(0, q, m) = (m-1)^q
$$
I conjecture that $R(n, 0, m)$ is a Chebyshev transform of $m^n$.
Examples of Chebyshev ...
- 4,948
1
vote
0
answers
68
views
On a numbers $k$ with specific $2$-adic valuation
Let $a(n)$ be A002326 (i.e., multiplicative order of $2 \operatorname{mod} 2n+1$).
Let $b(n)$ be A179382 (i.e., the smallest period of pseudo-arithmetic progression with initial term $1$ and ...
- 4,948
0
votes
0
answers
62
views
Linear recurrences in coefficients of powers of quotients of polynomial rings
It is known that linear recurrences with constant coefficients
can be computed via powers in $\mathbb{Z}[x]/f(x)$.
We believe that this generalizes to quotients of multivariate polynomial
rings.
Let $...
- 25.4k
3
votes
0
answers
69
views
Sequence that sum up to A343685
Let $a(n)$ be A343685 i.e.
$$
a(n)=2na(n-1)+\sum\limits_{j=0}^{n-1}\binom{n}{j}(n-j-1)!a(j), \\
a(0)=1
$$
Here the exponential generating function $A(x)$ satisfy
$$
A(x)=\frac{1}{1-2x+\log(1-x)}
$$
...
- 4,948
1
vote
0
answers
111
views
Recursion for the Bessel polynomial $y_n(x)$
Let $a(n)$ be A001515 i.e. the Bessel polynomial $y_n(x)$ evaluated at $x=1$. Here
$$
a(n) = (2n-1)a(n-1) + a(n-2), \\
a(0) = 1, a(1) = 2
$$
The closed form is
$$
a(n)=\sum\limits_{k=0}^{n}\binom{n+k}{...
- 4,948
1
vote
0
answers
94
views
Combinatorial interpretation for the more general case of $R(n,0)$
Let $f(n), g(n,m), h(n)$ be an arbitrary functions which equal to the non-negative integers.
Let
$$
R(n,q) = \sum\limits_{j=0}^{f(q)}g(q,j)R(n-1,j),\\
R(0,q) = h(q)
$$
In the comment to the one of ...
- 4,948
26
votes
1
answer
7k
views
Elegant recursion for A301897
Let $a(n)$ be A301897, i.e., number of permutations $b$ of length $n$ that satisfy the Diaconis-Graham inequality $I_n(b) + EX_n(b) \leqslant D_n(b)$ with equality. Here
$$a(n)=\frac{1}{n+1}\binom{2n}{...
- 4,948
2
votes
0
answers
126
views
Recurrence for A004208
Let $a(n)$ be A004208. Here
$$a(n)=n\prod\limits_{j=1}^{n}(2j-1)-\sum\limits_{i=1}^{n-1}a(i)\prod\limits_{j=1}^{n-i}(2j-1)$$
I conjecture that
$$a(n)=R(n-1,0)$$
where
$$R(n,q)=2(q+2)R(n-1,q+1)+\sum\...
- 4,948
3
votes
1
answer
140
views
Sequences that sum up to Dowling numbers
Let $a(n,k)$ be the sequence of $k$-Dowling numbers (for more information see A007405 and its CROSSREFS section) with e.g.f.
$$\operatorname{exp}\left(x + \frac{\operatorname{exp}(kx) - 1}{k}\right)$$
...
- 4,948
2
votes
0
answers
105
views
Sequences that sum up to the many sequences in the OEIS
Let
$$a(n,m,k)=\frac{1}{n}\sum\limits_{j=0}^{n}[n+kj\geqslant 0]\binom{n}{j}\binom{n+kj}{j-1}(m-1)^{j-1}$$
Here square brackets denote Iverson brackets.
There are many sequences in the OEIS that are ...
- 4,948
6
votes
1
answer
268
views
Sequence that sums up to the number of permutations avoiding the pattern $1-23-4$
Let $a(n)$ be A113227, i.e., the number of permutations on $[n]\equiv \{1, \ldots, n\}$ avoiding the pattern $1-23-4$.
The sequence begins with
$$1, 1, 2, 6, 23, 105, 549, 3207, 20577, 143239, 1071704,...
- 4,948
1
vote
0
answers
57
views
Recurrence for the number of permutations with a given excedance set
Let
$$\ell(n)=\left\lfloor\log_2 n\right\rfloor$$
$$f(n)=n-2^{\ell(n)}$$
$$T(n,k)=\left\lfloor\frac{n}{2^k}\right\rfloor\operatorname{mod}2$$
$$\operatorname{wt}(2n+1)=\operatorname{wt}(n)+1, \...
- 4,948
1
vote
0
answers
134
views
Recurrence for the A284005
Let
$$\ell(n)=\left\lfloor\log_2 n\right\rfloor$$
$$f(n)=n-2^{\ell(n)}$$
$$T(n,k)=\left\lfloor\frac{n}{2^k}\right\rfloor\operatorname{mod}2$$
$$\operatorname{wt}(2n+1)=\operatorname{wt}(n)+1, \...
- 4,948
0
votes
1
answer
101
views
Recurrence for the number of steps required to get one ball in each box
Given $n$ balls, all of which are initially in the first of $n$ numbered boxes, $a(n)$ is the number of steps required to get one ball in each box when a step consists of moving to the next box every ...
- 4,948
1
vote
0
answers
67
views
Recurrence for permutation of A007306 (denominators of Farey tree fractions)
Let $a(n)$ be A071585, i.e., numerator of the continued fraction expansion whose terms are the first-order differences of exponents in the binary representation of $4n$, with the exponents of $2$ ...
- 4,948
4
votes
2
answers
594
views
Squares in Lucas sequences
Good night, everyone!
According to a celebrated result by J. H. Cohn, the only perfect squares in the Fibonacci sequence are $F_{0}=0$, $F_{1}=F_{2}=1$, and $F_{12}=144$. It is also known that the ...
- 87
4
votes
1
answer
168
views
An inequality involving $k$-generalized Fibonacci numbers
I have worked on a Diophantine equation by using transcendental and reduction methods given by Baker and Davenport. However, to solve completely the equation I have one complicated case and I proved ...
- 41
7
votes
4
answers
1k
views
Find a formula for the recurrent sequence $q_{n+1}=q_n(q_n+1)+1$
Find an analytic formula for the recurrent sequence $$q_{n+1}=q_n(q_n+1)+1,\;\;q_0\in\mathbb N.$$
(The question was asked on 03.05.2018 by M. Pratsovytyi, see page 109 of Volume 1 of the Lviv ...
- 4,535
2
votes
1
answer
740
views
Power tower made of $2$s and $3$s: too high, too soon?
A power tower of a number $x$ is typified by
$$ x^{x^{x^{x^{x^{x^{x^{x^{x^x}}}}}}}}.$$
Here, however, we take the liberty of referring to the set $T$ of "$\{2,3\}$-power towers"; i.e., numbers
$$...
- 3,443
5
votes
1
answer
303
views
Simply generated sequences with mysterious differences
Suppose that $a_0 < a_1,$ $b_0 < b_1,$ and $$a_n=a_1b_{n-1}+a_0b_{n-2}+qn+r$$ for $n \geq 2$, where $a_0,a_1,b_0,b_1,q,r$ are integers such that $(a_n)$ and $(b_n)$ are increasing and ${(|a_n|)}$...
- 3,443
7
votes
2
answers
428
views
Limit associated with complementary sequences
Define $A=(a_n)$ and $B=(b_n)$ as follows: $a_0=1$, $a_1=2$, $b_0=3$, $b_1=4$, and $$a_n=a_0b_{n-1}+a_1b_{n-2}$$ for $n \geq 2$, where $A$ and $B$ are increasing and every positive integer occurs ...
- 3,443
1
vote
2
answers
534
views
Can these sequences stay integer-valued as many times as we want and then fail?
Edit:
Suppose that we choose some integer $d$ and some natural number $c=c_2$. Then if we plug those values into
$$ c_{n+1}=\frac{c_n(c_n+n+d)}n $$ and observe the behavior of this recursively ...
user114642
35
votes
0
answers
1k
views
Is there any positive integer sequence $c_{n+1}=\frac{c_n(c_n+n+d)}n$?
In a recent answer Max Alekseyev provided two recurrences of the form mentioned in the title which stay integer for a long time. However, they eventually fail.
QUESTION Is there any (added: ...
- 23.7k
9
votes
2
answers
1k
views
The p-adic valuation of a linear recurrence
Let $(u_n)_{n \geq 0}$ be an integer-valued linear recurrence of order $k \geq 1$. Precisely,
$$u_n = a_1 u_{n-1} + \cdots + a_k u_{n - k} \quad \forall n \geq k ,$$
for some $a_1, \ldots, a_k \in \...
user40023
1
vote
1
answer
276
views
Infinitely many sufficiently large powers in linear recurrences
Edit Aaron solved the original question with the
fourth order $$ a(n)=n2^n+\frac{(-1)^n-1^n}{2} $$
trying to make the question harder.
Let $a(n)$ be a linear recurrence with constant coefficients,
of ...
- 25.4k