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Results tagged with sequences-and-series
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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.
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
11
votes
Accepted
Is there a recurrence for the coefficients of the Laurent series expansion of $\frac{1}{1-e^...
$e^{e^x-1}$ is the exponential generating function for Bell numbers ${\cal B}_n$:
$$e^{e^x-1} = \sum_{n\geq 0} {\cal B}_n \frac{x^n}{n!}.$$
Then
$$g(x) := \frac{e^{e^x-1}-1}{x} = \sum_{n\geq 0} {\cal …
- 29.5k
10
votes
Binomial Coefficient Identity, Double Series, Floor Function
Notice that
$$\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
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
$$\sum_{n=1}^k \ …
- 29.5k
9
votes
Accepted
Proving a binomial sum identity
\begin{align}\sum_{n=0}^{\infty}\frac{\binom{2n+1}{n+1}}{2^{2n+1}\,(n+x+1)}&= \int_0^1\sum_{n=0}^{\infty}\frac{\binom{2n+2}{n+1}y^{n+x}}{2^{2n+2}}\,{\rm d}y\\&=\int_0^1 y^{x-1}\big((1-y)^{-1/2}-1\big) …
- 29.5k
8
votes
An explicit representation for polynomials generated by a power of $x/\sin(x)$
Faà di Bruno's formula implies that
$$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
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
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
$$b_n=\sum_{k=0}^n \binom{n}{2k}C_k \equiv\sum_{m\geq 0}\binom{n}{2(2^m-1)}\equiv 1+\nu_2(\l …
- 29.5k
7
votes
Accepted
An identity for product of central binomials
We have
$$\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
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
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
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
$$\frac1{k\binom{2k}{k}}=\frac12 B(k,k)=\frac12 \int_0^1 t^{k-1}(1-t)^{k-1}\,dt,$$
$$\frac{1}{k+1}+\cdots+\fra …
- 29.5k
6
votes
Accepted
Divisibility of (finite) power sum of integers
Notice that
$$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}.$$
It follo …
- 29.5k
6
votes
Convergence of $\sum(n^p\sin^qn)^{-1}$
A while ago I've addressed this question in On convergence of the Flint Hills series.
- 29.5k
6
votes
sum of odious numbers to the power of k
I think there is no simple formula here, although we can get some recurrence relations and related identities for generating functions as explained below.
Similarly to odious numbers, we have evil nu …
- 29.5k
6
votes
Accepted
An infinite series involving Jordan's totient function
The right hand side evaluates to
$$\sum_{m=0}^{\infty} m^{k-1} x^m,$$
and so it remains to verify that the coefficient of $x^m$ in the l.h.s. is also $m^{k-1}$.
By differentiating the l.h.s., we have
…
- 29.5k
5
votes
Accepted
Spherical Bessel functions. Sum of squares
From the definition 10.47.10, it follows that
$$\mathsf{j}_{n}^{2}(z)+\mathsf{y}_{n}^{2}(z) = h_n^{(1)}(z)\cdot h_n^{(2)}(z).$$
So, by the expansions 10.49.6 and 10.49.7,
\begin{split}
\mathsf{j}_{n}^ …
- 29.5k
5
votes
What is this sequence?
I've got that $a_{n,h}$ is the coefficient of $x^{n-1}$ in
\begin{split}
&(-1)^{n+h+1} \frac{n!}{2\cdot (h-1)!}\frac{\log(1+x)^{h}\left(\coth(-\frac{n}2\log(1+x))-1\right)}{1+x} \\
=\ &(-1)^{n+h} \fra …
- 29.5k
5
votes
Accepted
Number of positive integers $k$ such that there exists a nonnegative integer $m$ with $k + k...
The formula $c(n)=a(n+1)$ is pretty much straightforward, noticing that
$$\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.$$
The l …
- 29.5k
5
votes
Accepted
Sequences that sums up to second differences of Bell and Catalan numbers
Let me address the case of $s_2(n)$.
First we notice that $g(n-1) = k$ whenever $n=(2k+1)2^t$. Then
for $n=2^{t_1}(1+2^{1+t_2}(1+\dots(1+2^{1+t_\ell}))\dots)>1$ with $t_j\geq 0$, we have
$$a_2(n) = \b …
- 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
$$a_n-Q = 4u_nv_n - 2(2n^2-n+1)u_n - 2(2n^2-n+1)u_n - 8 …
- 29.5k
4
votes
Accepted
Series sum with coefficients that are Fibonacci numbers
First notice that
$$\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
4
votes
0
answers
274
views
Identities for powers of functions based on generalization of Lagrange interpolation
Lagrange polynomial can be used to obtain an identity:
$$(k+t)^n = \sum_{i=0}^n (k+d_i)^n \prod_{\substack{j=0\\ j\not=i}}^n \frac{t-d_j}{d_i-d_j},$$
which holds for any integer $n>0$, any real number …
- 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 $j\geq 1$
$$c_j = \left.\left(\frac{\partial}{\partial s}\right)^j \log( B(\alpha - e^s)(\beta-e^s) )\right|_{s=0} = \left. …
- 29.5k
4
votes
Accepted
Expanding in Fibonacci powers
Here is the direct proof without using the result of Ardila.
We will represent a sum of distinct Fibonacci numbers $m=\sum_{j=2}^n b_{j-1} F_j$, where $b_j\in\{0,1\}$, as a bit string $b_1b_2\ldots b_ …
- 29.5k
4
votes
Accepted
Why do convoluted convolved Fibonacci numbers pop up from this triangle?
We have $$T(n,k) = [x^ny^{n-k}]\ \frac{2 - (1+y)x}{1-(1+y)x-x^2}=[y^{n-k}]\ L_n(1+y),$$
where $L_n$ is the $n$-th Lucas polynomial.
For $k<n$, we have an explicit formula:
\begin{split}
T(n,k) &= \sum …
- 29.5k
4
votes
Identity involving double sum with binomials
Rewriting the l.h.s. of the conjectured identity and using the properties of beta function, we have:
\begin{split}
\text{l.h.s.} &= \sum_{B, b} (-1)^{a+b}{b\choose a} {B-1\choose A-1}\frac{1}{(B+b)\bi …
- 29.5k
4
votes
Accepted
Recurrence for the sum
Let $n=m^tk$ where $m\nmid k$. Then $f(n)=m^t$.
Furthermore, if $t>0$, then $f(n/m)=m^{t-1}$ and $n-f(n/m)=m^{t-1}(mk-1)$. It follows that $a(n)=a(m^{t-1}k)+a(m^{t-1}(mk-1))$ and further by induction …
- 29.5k
4
votes
Two conjectural series for $\pi$ involving the central trinomial coefficients
Not an answer, but a reduction to a definite integral.
First, Lagrange inversion implies that the generating function for $T_k$ is $${\cal T}(z):=(1-2z-3z^2)^{-\frac12}=((1+z)(1-3z))^{-\frac12},$$
a …
- 29.5k
3
votes
Summation of double exponential series
Notice that
$$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}.$$
The coefficients of powers of $q$ up to $q^{2^n-2}$ is …
- 29.5k
3
votes
Accepted
Yet another question about unrestricted partitions
I've established by brute-force that it is not possible to get a series with $\{-1,0,1\}$ coefficients this way. The maximum one can get is having such coefficients for degrees up to 121. Here is one …
- 29.5k
3
votes
The sequence $a(n)=(2^n \bmod p)^{p-1} \bmod p^2$
Using the notation from my answer to the previous question, we have
$$D(n+1)−(D(n+2)+1)\not\equiv 0\pmod{p}$$
if and only if $g_0(n+2) = 2 g_0(n+1) - p$ and $g_1(n+2) \equiv 2 g_1(n+1) + 1\pmod{p}$, i …
- 29.5k
3
votes
Accepted
Subsequences of odd powers
Quite similarly to my answer to the previous question, we have that for $n=2^tk$ with odd $k$,
$$
a(n)=\sum_{i=0}^t \binom{t}{i}p^{t-i}q^i a(2^i(k-1)+1).
$$
It further follows that for $n=2^{t_1}(1+2^ …
- 29.5k
3
votes
Accepted
Relation between $\sum_{k\ge 0}\binom {n+k}{k}a_k $ and $\sum_{k\ge 0}\binom {n+k}{k}\frac{a...
Since the second sum cannot start at $k=0$, I assume that both sums start at $k=1$.
Consider a particular example: $a_k = k\alpha^k$ with $|\alpha|<1$. Then
$$\sum_{k\geq 1} \binom{n+k}k \frac{a_k}k = …
- 29.5k
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
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
3
votes
Accepted
Matching two sequences between each other
A modification of Needleman–Wunsch algorithm for global alignment of $A$ and $B$, where you match/skip whole blocks (rather than individual symbols) in $B$ and do not allow gaps in $A$, will do the jo …
- 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
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
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
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:
$$\begin{ca …
- 29.5k
2
votes
Calculating "factorial sequence" of a rational function
Let's assume that $p$ is monic of degree $d$ and has distinct roots $r_1,\dots,r_d$. Consider the partial fraction decomposition:
$$f(x) = \frac{1}{p(x)} = \sum_{i=1}^d \frac{a_i}{x-r_i},$$
where $a_i …
- 29.5k
2
votes
Accepted
Sum with Stirling numbers of the second kind
The same idea of grouping terms by the number of unit bits, as well as grouping by the value of $\mathrm{wt}(j)$ (representing $j$ via individual bits) works here:
\begin{split}
s(n,m) &= \sum_{\ell=0 …
- 29.5k
2
votes
Guess (or upper bound) the general formula for a double sequence
First notice that $h(t)\geq t$ and $h(t+1)=3h(t)+1$ for all $t\geq 0$. Now, for $s>h(t+1)$, the recurrent formula reduces to
$$f(t+1,s) = f(t,s) + f(t,s-1),$$
which further implies that for all $t\geq …
- 29.5k
2
votes
Subsequence of the cubes
An explicit formula for $a(n)$ is derived in this answer. In particular, it gives
$$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
2
votes
Accepted
Formula from the recurrence relation
The conjectured formula can be proved by induction on $\mathrm{wt}(n)$.
For $\mathrm{wt}(n)=0$, we have $n=0$ and the conjectured formula trivially holds.
Now, for a positive integer $\ell$, suppose t …
- 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
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