Mats Granvik
• Member for 12 years, 8 months
• Last seen this week
Stats
7,326
reputation
167k
reached
63
197
questions
Communities
View all

Engineer.

$$T(n,k)$$ = https://oeis.org/A191898 $$(n,k)$$

$$\Lambda(n)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}$$ $$\sum\limits_{k=1}^{\infty}\sum\limits_{n=1}^{\infty} \frac{T(n,k)}{n^c \cdot k^s} = \sum\limits_{n=1}^{\infty} \frac{\lim\limits_{z \rightarrow s} \zeta(z)\sum\limits_{d|n} \frac{\mu(d)}{d^{(z-1)}}}{n^c} = \frac{\zeta(s) \zeta(c)}{\zeta(s + c - 1)}$$

$$-\frac{\zeta '(s)}{\zeta (s)}=\lim_{c\to 1} \, \left(\frac{\zeta (s) \zeta (c)}{\zeta (s+c-1)}-\zeta (c)\right)$$

$$\mu(n) = \underbrace{\underset{1 = n} 1 - \underset{a = n}{\sum_{a \geq 2}} 1 + \underset{ab = n}{\sum_{a \geq 2} \sum_{b \geq 2}} 1 - \underset{abc = n}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2}} 1 + \underset{abcd = n}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2} \sum_{d \geq 2}} 1 - \cdots}_{\text{#alternating sums}>\frac{\log(n)}{\log(2)}}$$

$$1/a^{b+i c}=1/a^b (\cos (c \log (1/a))+i \sin (c \log (1/a)))$$

1/a^(b + I*c) = 1/a^b*(Cos[c*Log[1/a]] + I*Sin[c*Log[1/a]])


$$f(n,s)=\frac{(s+1)^{n-1}+s-1}{s}$$

N[Table[2*Pi*Exp[1]*Exp[ProductLog[(n - 11/8)/Exp[1]]], {n, 1, 12}]]

Limit[(Zeta[s]*Zeta[s]/Zeta[2 s - 1] + Zeta[s]/Zeta[s - 1]), s -> 1]

Plot[RiemannSiegelTheta[t]/Pi +
Im[Log[Zeta[1/2 + I*t]] + I*Pi]/Pi, {t, 0, 60}, ImageSize -> Large]

Table[Limit[
Zeta[s] Total[1/Divisors[n]^(s - 1)*MoebiusMu[Divisors[n]]],
s -> 1], {n, 1, 32}]

x = N[Exp[-ZetaZero[1]/10], 100]
Sum[(-1)^k*x^(Log[k]*10), {k, 1, Infinity}]


$$\sum _{k=1}^{\infty } (-1)^{k+1} x^{c \log (k)}=0?$$ Keith Conrad: numbersoncircle.pdf

divided with: /2/PI()/EXP(1)

von Mangoldt function:

http://pastebin.com/u/MatsGranvik

Divisibility:

$$\mu(n) = \underbrace{\underset{1 = n} 1 - \underset{a = n}{\sum_{2 \leq a \leq \frac{n}{2^0} }} 1 + \underset{ab = n}{\sum_{2 \leq a \leq \frac{n}{2^1}} \sum_{2 \leq b \leq \frac{n}{2^1}}} 1 - \underset{abc = n}{\sum_{2 \leq a \leq \frac{n}{2^2}} \sum_{2 \leq b \leq \frac{n}{2^2}} \sum_{2 \leq c \leq \frac{n}{2^2}}} 1 + \underset{abcd = n}{\sum_{2 \leq a \leq \frac{n}{2^3}} \sum_{2 \leq b \leq \frac{n}{2^3}} \sum_{2 \leq c \leq \frac{n}{2^3}} \sum_{2 \leq d \leq \frac{n}{2^3}}} 1 - \cdots}_{\text{#alternating sums}>\frac{\log(n)}{\log(2)}}$$

3
34
66
24
Score
57
Posts
22
Posts %
21
Score
61
Posts
23
Posts %
11
Score
20
Posts
8
Posts %
9
Score
83
Posts
32
Posts %
7
Score
17
Posts
7
Posts %
2
Score
25
Posts
10
Posts %
Top posts
69
Jun 19, 2013
40
Feb 9, 2012
36
Jun 9, 2012
22
Jun 19, 2011
18
Apr 20, 2014
16
Mar 28, 2011
16
Jul 1, 2011
16
Jun 9, 2013
12
Sep 2, 2012
10
Sep 13, 2011
Top Meta posts
1
0
Top network posts
View all network posts