Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with sequences-and-series
Search options not deleted
user 7076
for questions about sequences and series, e.g. convergence, closed form expressions, etc. Note that there is a different tag for spectral sequences, and also note that MathOverflow is not for homework. Please consider consulting the online encyclopedia for integer sequences, if you are trying to identify a given sequence that you have found in your research.
6
votes
Accepted
Generalized Vieta-product
I doubt there exists a closed formula for $n\ne 2$. In the case $n=2$ such formula exists only thanks to the double-angle formula for cosine.
Let $n$ be fixed and $c=c_n$. Notice that $n=c^2-c$ and $ …
- 29.5k
2
votes
An elementary inequality for a recursive double sequence
So far I was not able to prove the inequality $\sigma_n(m)\leq\sigma_{n-1}(m)+m$ and have not even convinced myself that it is true. Still, I have deduced a number of properties that one may find help …
- 29.5k
1
vote
Accepted
upper bound involving recursions
Rewrite $a_{k+1}^2 \leq a_k^2 - a_k + \frac12$ as $4a_{k+1}^2 \leq (2a_k-1)^2 + 1$. Then
$$\frac{2a_{k+1}}{2a_k-1} \leq \left(1+\frac{1}{(2a_k-1)^2}\right)^{1/2} \leq 1 + \frac{1}{2(2a_k-1)^2} \leq 1 …
- 29.5k
2
votes
Calculating "factorial sequence" of a rational function
Let's assume that $p$ is monic of degree $d$ and has distinct roots $r_1,\dots,r_d$. Consider the partial fraction decomposition:
$$f(x) = \frac{1}{p(x)} = \sum_{i=1}^d \frac{a_i}{x-r_i},$$
where $a_i …
- 29.5k
3
votes
Closed form of $ \sum_{k_{j-1}=0}^{k_j}....\sum_{k_1=0}^{k_2} \sum_{k=0}^{k_1} k^m $ as a po...
So, the sum can be rewritten as
$$\sum_{0\leq k\leq k_1\leq\dots\leq k_j} k^m=\sum_{k=0}^{k_j} k^m \sum_{k\leq k_1\leq\dots\leq k_j} 1 = \sum_{k=0}^{k_j} k^m \binom{k_j-k+j-1}{j-1}$$
As explained in m …
- 29.5k
4
votes
Identity involving double sum with binomials
Rewriting the l.h.s. of the conjectured identity and using the properties of beta function, we have:
\begin{split}
\text{l.h.s.} &= \sum_{B, b} (-1)^{a+b}{b\choose a} {B-1\choose A-1}\frac{1}{(B+b)\bi …
- 29.5k
1
vote
Inverse problems for an asymptotic series which depends on a parameter?
Here is an approach that leads to an integral equation for
$$A(\nu,t) := \sum_{n=0}^{\infty}(-1)^{n}\frac{a_{n}(\nu)}{\nu} e^{Int}.$$
First we notice that
$$(-1)^{n}\frac{a_{n}(\nu)}{\nu} = \frac{1}{ …
- 29.5k
2
votes
Lower/Upper bounds for $ \sum\limits_{i=0}^k \binom ni x^i $
Let $p=\frac{x}{1+x}$ and $q=\frac{1}{1+x}$, and thus
$$\sum_{i=0}^k \binom{n}{i} x^i=(1+x)^n\sum_{i=n-k}^n \binom{n}{i} p^{n-i} q^i.$$
Then for $k<np$ Chernoff bound gives
$$\sum_{i=n-k}^n \binom{n …
- 29.5k
4
votes
is there any non prime (pseudoprime) that holds true for this test, or any prime that fall o...
Prime $p=7$ fails the test, but this is the only exception below $10^7$. This is likely explained by the characteristic polynomial $x^3 - x^2 - 2x + 1$ having discriminant $49=7^2$.
Also, there are n …
- 29.5k
4
votes
Is there any pseudoprime that pass this test above tested range, or any prime that does not ...
The sequence of pseudoprimes here starts with:
$$219781, 252601, 399001, 512461, 722261, 741751, 852841, 1024651, 1193221, 1533601, 1690501, 1735841, 1857241, 1909001, \dots$$
UPDATE. Efficient testi …
- 29.5k
4
votes
Accepted
A problem of divisibility
If $a_n\mid b_n$, then $a_n$ also divides
$$Q := (2n-1)a_n + 4b_n = (2n+1)^2 (u_n + v_n+4n).$$
We will blatantly require $a_n = Q$. Notice that
$$a_n-Q = 4u_nv_n - 2(2n^2-n+1)u_n - 2(2n^2-n+1)u_n - 8 …
- 29.5k
0
votes
For a given $n$, under what condition(s) there exists (at least) two different $c$ and $c′$ ...
It's easy to see that $X_n^c = X_n^{c+4}$ for any $n,c$. This implies that the set of suitable $n$ contains all integers $\geq 8$, for which we can take $c=0$ and $c'=4$.
Smaller $n$ can be tested m …
- 29.5k
10
votes
Binomial Coefficient Identity, Double Series, Floor Function
Notice that
$$\binom{n-i+\lfloor \frac{i}{2} \rfloor}{j}\binom{n-i+\lfloor \frac{i}{2} \rfloor-j}{\lfloor \frac{i}{2} \rfloor} = \binom{n-i}{j}\binom{n-i+\lfloor \frac{i}{2} \rfloor}{\lfloor \frac{i}{ …
- 29.5k
4
votes
Accepted
Series sum with coefficients that are Fibonacci numbers
First notice that
$$\frac{{m-1\choose{k}} {n-1\choose{k}}}{ {m+n-1\choose{2k+1}} {2k\choose{k}}} = \frac{(m-1)!(n-1)!(m+n-2-2k)!}{(2k+1)(n-1-k)!(m-1-k)!(m+n-1)!} = \frac{(2k+1)\binom{m+n-2-2k}{n-1- …
- 29.5k
3
votes
The sequence $a(n)=(2^n \bmod p)^{p-1} \bmod p^2$
Using the notation from my answer to the previous question, we have
$$D(n+1)−(D(n+2)+1)\not\equiv 0\pmod{p}$$
if and only if $g_0(n+2) = 2 g_0(n+1) - p$ and $g_1(n+2) \equiv 2 g_1(n+1) + 1\pmod{p}$, i …
- 29.5k
3
votes
Infinite sum of reciprocals of squares of lengths of tangents from origin to the curve $y=\s...
It needs to be pointed out that the series $\frac{\sin(x)-x\cos(x)}{x^3}$ comes from the MSE answer, and as I understand the claim there this series equals $\frac{1}{3}\prod_{k\geq 1} (1-\frac{x^2}{\l …
- 29.5k
8
votes
Accepted
A moment sequence and Motzkin numbers. Modular coincidence?
Let's start with $b_n$. Since Catalan number $C_k$ is odd iff $k=2^m-1$, from Lucas theorem it follows that
$$b_n=\sum_{k=0}^n \binom{n}{2k}C_k \equiv\sum_{m\geq 0}\binom{n}{2(2^m-1)}\equiv 1+\nu_2(\l …
- 29.5k
8
votes
Accepted
Is Somos-8 $\mod 2$ periodic?
I confirm the observation of @მამუკაჯიბლაძე that $\nu_2(s_{103})=-1$. That is, $s_n\bmod 2$ is not well-defined at first place, which invalidates the question.
Moreover, for $n\geq 133$, $\nu_2(s_n)$ …
- 29.5k
1
vote
Upper bound over $[0,1] $ for strange family of polynomials
We can get the generating function for $P_n(x)$ and prove the property $P_n(x)=(-1)^{n+1}P_n(1-x)$ as follows.
Noticing that $\sum_p (-1)^{p+1} \frac{z^p}{p} = \log(1+z)$, we conclude that
$$a_{n,k} …
- 29.5k
3
votes
Accepted
Closed formula for reversion of Jacobi theta series
Perhaps, the form given by Lagrange inversion theorem cannot be much simplified here. It expresses the $n$-th coefficient of a series reversion as the sum of $n-1$ values of exponential Bell polynomia …
- 29.5k
5
votes
Accepted
Number of positive integers $k$ such that there exists a nonnegative integer $m$ with $k + k...
The formula $c(n)=a(n+1)$ is pretty much straightforward, noticing that
$$\lfloor \log_{i+1}(n-i)\rfloor - \lfloor\log_{i+1}(n-1-i)\rfloor=1\quad\text{iff}\quad n-i=(i+1)^m\text{ for some }m.$$
The l …
- 29.5k
6
votes
Convergence of $\sum(n^p\sin^qn)^{-1}$
A while ago I've addressed this question in On convergence of the Flint Hills series.
- 29.5k
2
votes
Accepted
Sequence that sums up to INVERTi transform applied to the ordered Bell numbers
First, we let $P(j,k):=1+f(\lfloor\tfrac{j}{2^k}\rfloor+1)$ and sum over $n$ of fixed weight $\ell:=\mathrm{wt}(n)$ (like in this answer):
\begin{split}
s(n) &= \sum_{\ell=0}^n \sum_{t_1 + \dots + t_\ …
- 29.5k
3
votes
Number of numbers in $n$th difference sequence
It's easy to see that $D_n\leq \texttt{A029579}(n)$. Indeed, $\Delta^1$ is a Sturmian word, which is known to have exactly $n+1$ factors of length $n$.
Now, $\Delta^n$ is formed by values of the $(n …
- 29.5k
1
vote
Multivariate generating function
The given g.f. directly follows from the sequence definition.
Indeed, the sequence gives the number of partitions of vector $(n,m)$ into the sum of vectors with nonnegative integer components. Each p …
- 29.5k
6
votes
sum of odious numbers to the power of k
I think there is no simple formula here, although we can get some recurrence relations and related identities for generating functions as explained below.
Similarly to odious numbers, we have evil nu …
- 29.5k
8
votes
An explicit representation for polynomials generated by a power of $x/\sin(x)$
Faà di Bruno's formula implies that
$$d_k(n) =
\sum_{m_1,\dots,m_k\geq 0\atop 1\cdot m_1+\cdots+k\cdot m_k=k} \frac{(2k)!}{m_1!\,2!^{m_1}\,m_2!\,4!^{m_2}\,\cdots\,m_k!\,(2k)!^{m_k}}\cdot (n)_{m_1+\cd …
- 29.5k
3
votes
Summation of double exponential series
Notice that
$$S(q,n) = \sum_{i=1}^n q^{2^i} = q^2\sum_{i=1}^n q^{2^i-2}$$
and thus
$$p_n(q) \approx \left( \sum_{i=1}^n q^{2^i-2}\right)^{-1}.$$
The coefficients of powers of $q$ up to $q^{2^n-2}$ is …
- 29.5k
3
votes
Accepted
Yet another question about unrestricted partitions
I've established by brute-force that it is not possible to get a series with $\{-1,0,1\}$ coefficients this way. The maximum one can get is having such coefficients for degrees up to 121. Here is one …
- 29.5k
1
vote
Accepted
Expansion of inverse logarithmic integral in terms of lambert w
Let me elaborate on reuns' idea and show how one can find coefficients $a_i$ by solving a certain ODE.
Let's define:
$$f(z) := \sum_{i\geq 3} a_i z^{i-3}$$
so that we get a functional equation:
$$\mat …
- 29.5k
0
votes
Prove a family of series having integer coefficients
As a follow-up to js21's answer, the following explicit formula for $F_r(x^2)$ holds:
$$F_r(x^2) = \int_0^{\infty} \cos(xt)^r e^{-t} dt.$$
- 29.5k
1
vote
Accepted
Is this recurrent sequence decreasing?
Since $x_0=0$, it will be convenient to do summation starting from $t=0$.
Denoting $r:=1-p-q$, we have
\begin{split}
S_n &= \frac1n\sum_{t=0}^n \left(pr^{2t} + (q-p)r^{t} - q\right) \\
&=\frac{p}{1-r^ …
- 29.5k
4
votes
Accepted
aproximate sum involving binomial coefficients
Just a rough idea. Let $\alpha, \beta$ be the zeros of $1+Ax+Bx^2$, then for $j\geq 1$
$$c_j = \left.\left(\frac{\partial}{\partial s}\right)^j \log( B(\alpha - e^s)(\beta-e^s) )\right|_{s=0} = \left. …
- 29.5k
3
votes
Accepted
Relation between $\sum_{k\ge 0}\binom {n+k}{k}a_k $ and $\sum_{k\ge 0}\binom {n+k}{k}\frac{a...
Since the second sum cannot start at $k=0$, I assume that both sums start at $k=1$.
Consider a particular example: $a_k = k\alpha^k$ with $|\alpha|<1$. Then
$$\sum_{k\geq 1} \binom{n+k}k \frac{a_k}k = …
- 29.5k
2
votes
Accepted
Sum with Stirling numbers of the second kind
The same idea of grouping terms by the number of unit bits, as well as grouping by the value of $\mathrm{wt}(j)$ (representing $j$ via individual bits) works here:
\begin{split}
s(n,m) &= \sum_{\ell=0 …
- 29.5k
1
vote
Prove that when converge, the following expansions are equal
The sum in the denominator of $f_3(x)$ evaluates to
$$\frac{(-1)^n}{(x-n)\binom{x}{n}n!}.$$
Hence, the equality $f_2(x) = f_3(x)$ is straightforward.
- 29.5k
6
votes
Evaluations of three series involving binomial coefficients
Here is a proof of the first identity. The others can probably be done similarly.
We have
$$\frac1{k\binom{2k}{k}}=\frac12 B(k,k)=\frac12 \int_0^1 t^{k-1}(1-t)^{k-1}\,dt,$$
$$\frac{1}{k+1}+\cdots+\fra …
- 29.5k
10
votes
Accepted
The square root of natural number expressed by an infinite series
So, here we have $P=2$ and $Q=1-m$.
Notice that
$$\frac{Q^n}{U_n(P,Q)U_{n+1}(P,Q)} = \frac{U_{n+1}(P,Q)}{U_n(P,Q)}-\frac{U_{n+2}(P,Q)}{U_{n+1}(P,Q)}.$$
By telescoping, it follows that
$$\sum_{n=1}^k \ …
- 29.5k
6
votes
Accepted
Divisibility of (finite) power sum of integers
Notice that
$$2S_a(b) \equiv 1^{2b} + 2^{2b} + \dots + (6a-4)^{2b} \pmod{6a-3}.$$
From Faulhaber's formula, we have
$$1^{2b} + 2^{2b} + \dots + (6a-4)^{2b} \equiv B_{2b} (6a-3) \pmod{6a-3}.$$
It follo …
- 29.5k
17
votes
Accepted
2-adic valuation of a certain binomial sum
First we notice that
\begin{split}
a_n & = n \int_0^1 x^n (1+x)^{n-1}{\rm d}x \\
& = n \int_0^1 (1-x)^n (2-x)^{n-1}{\rm d}x \\
& = n\sum_{k=0}^{n-1} \binom{n-1}{k}2^k (-1)^{n-1-k} \int_0^1 (1-x)^n x^ …
- 29.5k
4
votes
Accepted
Recurrence for the sum
Let $n=m^tk$ where $m\nmid k$. Then $f(n)=m^t$.
Furthermore, if $t>0$, then $f(n/m)=m^{t-1}$ and $n-f(n/m)=m^{t-1}(mk-1)$. It follows that $a(n)=a(m^{t-1}k)+a(m^{t-1}(mk-1))$ and further by induction …
- 29.5k
7
votes
Attempt at applying linear programming to the partial sums of the Möbius inverse of the Harm...
Here is a proof of Conjecture 2.
First, we have
\begin{split}
\sum_{k=1}^n M(n,k) &= \sum_{k=1}^n \sum_{m=k}^n \sum_{d|\gcd(m,k)} d\cdot\mu(d) \\
&= \sum_{m=1}^n \sum_{k=1}^m \sum_{d|\gcd(m,k)} d\cdo …
- 29.5k
6
votes
Accepted
An infinite series involving Jordan's totient function
The right hand side evaluates to
$$\sum_{m=0}^{\infty} m^{k-1} x^m,$$
and so it remains to verify that the coefficient of $x^m$ in the l.h.s. is also $m^{k-1}$.
By differentiating the l.h.s., we have
…
- 29.5k
0
votes
Accepted
Generating function for partial sums of the sequence
As proved in this answer, if we represent $n$ as $n=2^{t_1}(1+2^{t_2+1}(1+\dots(1+2^{t_m+1}))\dots)$ with $t_j\geq 0$, then
\begin{split}
a(n) = P(\ell)^{t_1}\prod_{j=1}^{\ell-1} P(\ell-j)^{t_{2j}+t_{ …
- 29.5k
4
votes
Two conjectural series for $\pi$ involving the central trinomial coefficients
Not an answer, but a reduction to a definite integral.
First, Lagrange inversion implies that the generating function for $T_k$ is $${\cal T}(z):=(1-2z-3z^2)^{-\frac12}=((1+z)(1-3z))^{-\frac12},$$
a …
- 29.5k
2
votes
Two-term recurrence relation
I assume that the initial conditions $a_0,a_1,b_0,b_1$ and that $n\to +\infty$.
Let $A(x):=\sum_{n\geq 0} a_n x^n$ and $B(x):=\sum_{n\geq 0} a_n x^n$. Then the recurrence relations become:
$$\begin{ca …
- 29.5k
7
votes
Accepted
An identity for product of central binomials
We have
$$\prod_{j=1}^n \binom{2j}{j} = \frac{2!4!\cdots (2n)!}{(1!2!\cdots n!)^2}$$
and
$$\prod_{j=1}^n 2\binom{n+j}{2j} = 2^n\frac{(n+1)!(n+2)!\cdots (2n)!}{(2!4!\cdots (2n)!)\cdot (0!1!\cdots (n-1) …
- 29.5k
4
votes
Accepted
Expanding in Fibonacci powers
Here is the direct proof without using the result of Ardila.
We will represent a sum of distinct Fibonacci numbers $m=\sum_{j=2}^n b_{j-1} F_j$, where $b_j\in\{0,1\}$, as a bit string $b_1b_2\ldots b_ …
- 29.5k
2
votes
Subsequence of the cubes
An explicit formula for $a(n)$ is derived in this answer. In particular, it gives
$$a(\tfrac{4^n-1}{3})=\left[\prod_{j=1}^{\lfloor (n-1)/2\rfloor} \bigg(q^{\lfloor (n+1)/2\rfloor-j}+p\tfrac{q^{\lfloor …
- 29.5k
0
votes
Proof that $3^ns + \sum_{k=0}^{n-1} 3^{n-k-1}2^{a_k}=2^m.$
I found it interesting to consider the problem with the minus rather than plus sign in front of the sum:
$$(\star)\qquad 3^n s - \sum_{k=0}^{n-1} 3^{n-1-k} 2^{a_k} = 2^m.$$
Consider the function
$$f …
- 29.5k