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7h
accepted Logarithmically bounded function fulfills $f(n) \le \lceil m \cdot \log_b r \rceil$ for certain numbers $n,m,r$
7h
comment Logarithmically bounded function fulfills $f(n) \le \lceil m \cdot \log_b r \rceil$ for certain numbers $n,m,r$
Ok, I see your example even works for $r = 2$, so it shows that even under these stronger assumption the inequality does not hold.
7h
comment Logarithmically bounded function fulfills $f(n) \le \lceil m \cdot \log_b r \rceil$ for certain numbers $n,m,r$
And the assumption $f(n) = 1 + \log_2(n)$ is also now valid, as $f(n) \in \mathbb N$ must hold.
7h
revised Logarithmically bounded function fulfills $f(n) \le \lceil m \cdot \log_b r \rceil$ for certain numbers $n,m,r$
added 51 characters in body
7h
comment Logarithmically bounded function fulfills $f(n) \le \lceil m \cdot \log_b r \rceil$ for certain numbers $n,m,r$
Oh, I am sorry, I meant all values to be integers, so $r \in \mathbb N$ and $r = 1.1$ would not be allowed. I will edit my post..
7h
revised Logarithmically bounded function fulfills $f(n) \le \lceil m \cdot \log_b r \rceil$ for certain numbers $n,m,r$
added 12 characters in body
7h
comment Logarithmically bounded function fulfills $f(n) \le \lceil m \cdot \log_b r \rceil$ for certain numbers $n,m,r$
Yes! I will edit my post.
11h
revised How do I efficiently budget for digits when converting from one base to another?
Saw a error in the bounds and their relation to the number of digits by one unit
12h
asked Logarithmically bounded function fulfills $f(n) \le \lceil m \cdot \log_b r \rceil$ for certain numbers $n,m,r$
16h
accepted Intuition on Martin-Löf-Test for finite strings
2d
answered How do I efficiently budget for digits when converting from one base to another?
2d
accepted Show $n \cdot \log_b r + \log_b \frac{r}{r-1} \le \lceil n \cdot \log_b r \rceil$
2d
asked Show $n \cdot \log_b r + \log_b \frac{r}{r-1} \le \lceil n \cdot \log_b r \rceil$
Jul
31
revised Representing numbers by quasilexicographic ordered strings, formula for size of conversion between different alphabets
edited tags
Jul
31
asked Representing numbers by quasilexicographic ordered strings, formula for size of conversion between different alphabets
Jul
31
accepted Order-preserving map between $X^*$ and $\mathbb N$ for some finite set $X$
Jul
31
asked Order-preserving map between $X^*$ and $\mathbb N$ for some finite set $X$
Jul
31
comment A simple expression to map $\mathbb N^*$ bijectively to $\mathbb N$
@J.E.-Pin: In my post I wrote $\mathbb N = \{1,2,3,\ldots\}$ and $\mathbb N_0 = \mathbb N \cup \{0\}$. But indeed that doesn't matter that much conceptually, otherwise $(n_1,\ldots, n_k) \mapsto 10^{n_1-1}10^{n_2-1}\cdots 10^{n_k-1}$ would work too. Also note that I do not included the empty word in the union $\bigcup_{k=1}^{\infty} \mathbb N^k$, I know this is not in concordance with standard practice in string theory/computer science, so sorry, but again this does not change anything conceputally much.
Jul
31
comment A simple expression to map $\mathbb N^*$ bijectively to $\mathbb N$
@RobArthan: $f((0,0)) = 11, f((0) = 1, f(()) = 0$, so this is fine. What he does is coding each tupel by representing each number unary (i.e. $n \leftrightarrow 0^n$) and using $1$ as a separating sign, and in that way each binary digit/sequence (without leading zeros) occurs too, giving a surjection. In retrospect a really simple idea :)
Jul
31
comment Proof of a classical Theorem of Martin-Löf on complexity dips for Kolmogorov complexity,
Yes, for fixed $x_1\cdots x_m$. I see in the construction that by selecting some fixed $x_1\cdots x_m$ and then always appending combinations, that this occurs in infinitely many $A_n$'s, but I see not that all occur that way. For example consider $1,11,111,\ldots$ and the intervals they describe, then for each $n$ the string $1^n$ occurs i.o. but that does not mean that each string $w$ occurs infintely often (and in fact it does not in this example, if some $0$ occurs in $w$ then it occurs just finitely often). Also why $\sum_{i=n}^{\infty} \mu(A_n) = \infty$ (this point may be crucial...)?