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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.
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
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
$$g(x) := \frac{e^{e^x-1}-1}{x} = \sum_{n\geq 0} {\cal …
- 29.5k
2
votes
Guess (or upper bound) the general formula for a double sequence
First notice that $h(t)\geq t$ and $h(t+1)=3h(t)+1$ for all $t\geq 0$. Now, for $s>h(t+1)$, the recurrent formula reduces to
$$f(t+1,s) = f(t,s) + f(t,s-1),$$
which further implies that for all $t\geq …
- 29.5k
3
votes
Accepted
Matching two sequences between each other
A modification of Needleman–Wunsch algorithm for global alignment of $A$ and $B$, where you match/skip whole blocks (rather than individual symbols) in $B$ and do not allow gaps in $A$, will do the jo …
- 29.5k
1
vote
Accepted
Pair of recurrence relations with $a(2n+1)=a(2f(n))$
As proved in this answer, for $n=2^tk$ with $2\nmid k$, we have
$$a_1(n)=\sum_{i=0}^t \binom{t}{i} a_1(2^i(k-1)+1).$$
Then for $n=2^{t_1}(1+2^{t_2}(1+\dots(1+2^{t_m}))\dots)$ with $t_1\geq 0$ and $t_j …
- 29.5k
5
votes
Accepted
Spherical Bessel functions. Sum of squares
From the definition 10.47.10, it follows that
$$\mathsf{j}_{n}^{2}(z)+\mathsf{y}_{n}^{2}(z) = h_n^{(1)}(z)\cdot h_n^{(2)}(z).$$
So, by the expansions 10.49.6 and 10.49.7,
\begin{split}
\mathsf{j}_{n}^ …
- 29.5k
5
votes
What is this sequence?
I've got that $a_{n,h}$ is the coefficient of $x^{n-1}$ in
\begin{split}
&(-1)^{n+h+1} \frac{n!}{2\cdot (h-1)!}\frac{\log(1+x)^{h}\left(\coth(-\frac{n}2\log(1+x))-1\right)}{1+x} \\
=\ &(-1)^{n+h} \fra …
- 29.5k
9
votes
Accepted
Proving a binomial sum identity
\begin{align}\sum_{n=0}^{\infty}\frac{\binom{2n+1}{n+1}}{2^{2n+1}\,(n+x+1)}&= \int_0^1\sum_{n=0}^{\infty}\frac{\binom{2n+2}{n+1}y^{n+x}}{2^{2n+2}}\,{\rm d}y\\&=\int_0^1 y^{x-1}\big((1-y)^{-1/2}-1\big) …
- 29.5k
2
votes
Accepted
Formula from the recurrence relation
The conjectured formula can be proved by induction on $\mathrm{wt}(n)$.
For $\mathrm{wt}(n)=0$, we have $n=0$ and the conjectured formula trivially holds.
Now, for a positive integer $\ell$, suppose t …
- 29.5k
4
votes
Accepted
Why do convoluted convolved Fibonacci numbers pop up from this triangle?
We have $$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),$$
where $L_n$ is the $n$-th Lucas polynomial.
For $k<n$, we have an explicit formula:
\begin{split}
T(n,k) &= \sum …
- 29.5k
1
vote
Accepted
Simplification of $\sum_{m=0}^\infty \text B_z(m+a,b-m)x^m,\sum_{m=0}^\infty \frac{\text B_z...
I'm not sure if this is considered simpler, but still:
\begin{split}
\sum_{m=0}^\infty \frac{\text B_z(m+a,b-m)x^m}{m!} &= \sum_{m=0}^\infty \frac{x^m}{m!}\int_0^z y^{m+a-1}(1-y)^{b-m-1}\,{\rm d}y\\
& …
- 29.5k
4
votes
0
answers
274
views
Identities for powers of functions based on generalization of Lagrange interpolation
Lagrange polynomial can be used to obtain an identity:
$$(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},$$
which holds for any integer $n>0$, any real number …
- 29.5k
2
votes
Accepted
Partition of $(2^{n+1}+1)2^{2^{n-1}+n-1}-1$ into parts with binary weight equals $2^{n-1}+n$
Notice that for $i\in\{0,1,\dots,2^{n-1}+n\}$ we have
$$a(i+1,2^{n-1}+n) = 2^{2^{n-1}+n+1} - 1 - 2^{2^{n-1}+n-i}.$$
Then the sum in question can be easily computed:
\begin{split}
& a(1,2^{n-1}+n)+\sum …
- 29.5k
5
votes
Accepted
Sequences that sums up to second differences of Bell and Catalan numbers
Let me address the case of $s_2(n)$.
First we notice that $g(n-1) = k$ whenever $n=(2k+1)2^t$. Then
for $n=2^{t_1}(1+2^{1+t_2}(1+\dots(1+2^{1+t_\ell}))\dots)>1$ with $t_j\geq 0$, we have
$$a_2(n) = \b …
- 29.5k
2
votes
Accepted
Coefficients of number of the same terms which are arising from iterations based on binary e...
In other words, if $(b_\ell b_{\ell-1}\dots b_0)_2$ is the binary representation of $n$, then
$$a(n) = g(g(\dots g(g(0,b_0),b_1)\dots ),b_{\ell-1}), b_\ell),$$
where
$$g(A,b) = \begin{cases} A+2, &\te …
- 29.5k