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Results tagged with nt.number-theory
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user 7076
Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
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
$$g(A,b) = \begin{cases} A+2, &\te …
- 29.5k
2
votes
Accepted
Partition of $(2^{n+1}+1)2^{2^{n-1}+n-1}-1$ into parts with binary weight equals $2^{n-1}+n$
Notice that for $i\in\{0,1,\dots,2^{n-1}+n\}$ we have
$$a(i+1,2^{n-1}+n) = 2^{2^{n-1}+n+1} - 1 - 2^{2^{n-1}+n-i}.$$
Then the sum in question can be easily computed:
\begin{split}
& a(1,2^{n-1}+n)+\sum …
- 29.5k
6
votes
Request for an exact formula related to a partition in number theory
$F(a_1,\dots,a_n;b)$ equals the coefficient of $z^b$ in the generating function
$$f(z):=\frac{1}{1-z^{a_1}}\frac{1}{1-z^{a_2}}\cdots \frac{1}{1-z^{a_n}}.$$
For a fixed choice of $a_1,\dots,a_n$, expl …
- 29.5k
2
votes
Existence of solution for a system of quadratic diophantine equations / symmetric quadratic ...
UPDATE. Using factorization $-2(x+1)^2x^{p-3}$ over the corresponding number field, I established that there are no solutions for $p=17$. Furthermore, I computationally verified that for primes $p<30$ …
- 29.5k
2
votes
Calculate the great common factor between $2^{2n+1}-1$ and $2^{4m+2}+1$
In general we have
$$\gcd(2^a + 1, 2^b - 1) = \frac{2^{\gcd(2a,b)}-1}{2^{\gcd(a,b)}-1},$$
which is evaluates to $1$ for $a=4m+2$ and $b=2n+1$.
- 29.5k
10
votes
Accepted
Ordinary partitions vs partitions into odd parts
The g.f. for the right-hand is
$$e^{2x}\cdot e^{2x^2} \cdots = e^{2\frac{x}{1-x}}.$$
For the left-hand side, additionally introducing variable $y$ to account for partition length, we get
$$\sum_{n\ge …
- 29.5k
3
votes
Accepted
Inductively computing Mersenne primes / perfect numbers?
The conjecture fails for $n=8128$, which can be verified in matter of seconds as explained below. I used PARI/GP for my verification.
First, since the conjecture concerns only values of at $x$'s being …
- 29.5k
8
votes
Accepted
Is there a similar formula like Ramanunjan's Eisenstein series identity for $\sum_{k=1}^{n-1...
Numerical experiments suggest that
$$A_2(n) := \sum_{k=1}^{n-1} k^2\sigma(k)\sigma(n-k) = \frac{n^2}{8}\sigma_3(n) - \frac{4n^3-n^2}{24}\sigma(n).$$
PS. In fact, it directly follows from the quoted To …
- 29.5k