User blog:Wythagoras/Results of the First International Googological Olympiad

The First International Googological Olympiad has ended. Here will follow the results, remarks and solutions. In the First International Googological Olympiad (FIGO for short) have participated four contestants, of which three have sent all problems to me. I consider the olympiad as a succes, given the fact that all contestants responded very positive. I've learned a lot of this olympiad, and I've enjoyed organizing it.

Results
* These scores are 0.0 because I haven't received these problems.

Problems
'''Problem 1. '''Find a number \(n\) such that \(\frac{\sigma(n)}{n}\) is larger than 1,000. (\(\sigma(n)\) is the sum of the divisors of \(n\)).

'''Problem 2. '''Proof that \(SCG(2n)\) is even for every \(n \in \mathbb N\). (empty graph is counted)

'''Problem 3. '''Determine all quadruples of numbers (a,b,c,d) such that

\[a \rightarrow b \rightarrow c \rightarrow d > \{a,b,c,d\}\]

'''Problem 4. '''A function f is k-exponential iff f(n) = E(Θ(n))#k, using Hyper-E. For example, a function f is 3-exponential iff \(f(n) = 2^{2^{2^{\Theta(n)}}}\). Find, for every \(k \in \mathbb N\), a function f such that \(f^n(n)\) is a k-exponential function. Give the function in terms of: addition, multiplication, exponentation, factorial, logarithm, iterated logarithm, tetration, subtraction and division.

Solutions
Problem 1. Find a number \(n\) such that \(\frac{\sigma(n)}{n}\) is larger than 1,000. (\(\sigma(n)\) is the sum of the divisors of \(n\)).

Solution. We'll show that \(N = \text{ceil}(e^{1000})\#\) statistifies. (# is the primorial function).

\(N\) is divisible by all primes up to \(\text{ceil}(e^{1000})\). Therefore it is also divisble by numbers up to \(\text{ceil}(e^{1000})\), because they can be factorized into primes less than \(\text{ceil}(e^{1000})\).Therefore \(N\) is divisible by all numbers in the form \(\frac{N}{k}\), with \(1 \leq k \leq \text{ceil}(e^{1000})\). Therefore:

\[\frac{\sigma(N)}{N} > \sum^{\text{ceil}(e^{1000})}_{k=0} \frac{1}{k} > ln(\text{ceil}(e^{1000})) > ln(e^{1000}) = 1000\]

Conclusion: \(N = \text{ceil}(e^{1000})\#\) statistifies the given condition. q.e.d.

Problem 2. Proof that \(SCG(2n)\) is even for every \(n \in \mathbb N\). (empty graph is counted)

Solution. We'll show that the sequence consists of two parts: one part with and one part without edges. Suppose \(S\) is the longest sequence of graphs and it has a graph with edges after a graph. Call \(G\) the first graph without edges, \(v\) the number of vertices of this graph and \(H\) the graph with edges after it. \(H\) is by definition not homeomorphically embeddable into \(G\) or earlier graphs. However, any graph without edges is not homeomorphically embeddable into a graph with edges. Therefore we can have \(H\) at the place of \(G\) and then a graph with \(v+1\) vertices, then \(G\), and then the original sequence with \(H\) removed. This sequence is, however, longer. Contradiction.

Therefore the sequence consisting out of one part with and one part without edges is the longest. Let \(G_n\) be the last graph with a vertex for \(SCG(2i)\). It is optimal when the first graph without edges has as many vertices as possible, in this case \(n+2i+1\). Further, it is optimal when each graph has only one vertex less than the graph before it. Therefore, this results in a sequence of \(n+2i+2\) graphs without edges, and therefore a total of \(2n+2i+2\) graphs. Because this is divisible by two, the proof is complete. q.e.d.

'''Problem 3. '''Determine all quadruples of numbers (a,b,c,d) such that

\[a \rightarrow b \rightarrow c \rightarrow d > \{a,b,c,d\}\]

'''Solution. '''When a=b=2, we know that both the chained arrow expression and the BEAF expression are equal to 4. When a=1, both expressions are equal to 1. When b=1, both expressions are equal to a. When d=1, both expressions are equal to \(a \uparrow^c b\). Therefore, in these cases, there is no strict inequality.

Then, when a≥3, b≥2, c≥1 and d≥2. We'll use and prove a variant of Bird's Proof. We want to proof that :

\[\forall a\geq3, b\geq2, c\geq1, d\geq2: \{a,b,c,d\} \geq a \rightarrow b \rightarrow c \rightarrow d\]

First, some lemmas: 

''L1. ''a^b ≥ 3^b ≥ b+2. for a≥3,b≥2. Proof: By Bernoulli's inequality, (2+1)^b ≥ 2b+1 ≥ b+3 > b+2.

''L2. ''\(a\geq3,b\geq2: a\uparrow^b a \geq a^b+1 > a^b\). Proof by induction. Base case: b=2: We know that \(a \uparrow^2 a > a^a > a^2\). Step: \(a \uparrow^b a > a \uparrow^{b-1} (a \uparrow^{b-1} a) > a \uparrow^{b-1} 2a\). Now, by the Knuth Arrow Theorem, \(a \uparrow^{b-1} 2a > (a \uparrow^{b-1} a) \uparrow^{b-1} a\). By the induction hypothesis is this larger than \((a^{b-1}) \uparrow^{b-1} a > (a^{b-1})^(a^{b-1}) \geq (a^{b-1})^3 > a^{3b-3} \geq a^{b+1} > a^b\). Because we work with integers, this also shows that \(a\uparrow^b a \geq a^b+1\). This completes the proof.

Now, the main proof. We induction on \(c\).

''Base. ''Now, we prove that the inequality holds for a≥3, b≥2, c=1 and d≥2 by induction. Chained arrow returns \(a^b\). When b=2, the inequality states \(\{a,2,1,d\} \geq \{a,2,1,2\} = \{a,a,a\} > a^2\), which is correct (because a≥3).

For the induction step: \(\{a,b+1,1,d\} \geq \{a,b+1,1,2\} = \{a,a,\{a,b,1,2\}\} >^{IH} \{a,a,a^b\} \geq \{a,a,b+3\} > a^{b+1} \). First inequality follows form the IH, second fom L1, third form L2. This completes the induction.

''Step. ''Now, we prove that if it is valid for all (a,b,c,d) if it is true for (a,b,c-1,d) which follows form the induction hypothesis.

\[a \rightarrow b \rightarrow c \rightarrow d = a \rightarrow b \rightarrow (a \rightarrow b \rightarrow (c-1) \rightarrow d) \rightarrow (d-1) \]

\[< \{a,b,\{a,b,c-1,d\},d-1\} < \{a,a^{b-1},\{a,a^{b-1},c-1,d\},d-1\} = \{a,a^{b-1},c-1,d\} \geq \{a,\{a,b-1,c,d\},c-1,d\} = \{a,b,c,d\}\]

First inequality follows form the IH, second fom L1, third from the properties of BEAF. This completes the induction.

The only case that remains is a=2, b≥3, c≥1, d≥2. Here, BEAF returns 4 as a result of Cookiefonster, while the chained arrow notation returns a number larger than or equal to \(2^3=8\). Therefore all solutions have the form of:

\[(a,b,c,d) \text{with} a=2, b\geq3, c\geq1, d\geq2\]

q.e.d.

Problem 4. A function f is k-exponential iff f(n) = E(Θ(n))#k, using Hyper-E. For example, a function f is 3-exponential iff \(f(n) = 2^{2^{2^{\Theta(n)}}}\). Find, for every \(k \in \mathbb N\), a function f such that \(f^n(n)\) (function iteration) is a k-exponential function. Give the function in terms of: addition, multiplication, exponentation, factorial, logarithm, iterated logarithm, tetration, subtraction and division.

'''Solution. '''The following set of functions statistifies the problem:

\[f(n) = \text{exp}^{k-2}(\text{ln}^{k-2}(n)^2) \]

Here the exponent is function iteration, and taking inverse ufncitons when it is negative.

When k=1, the function is equal to 2n. When iterated, this becomes \(2^nn\), which is one-exponential.

Further, when k≥2, we give a proof by induction to \(n\) that \(f^n(r)\) equals \[\text{exp}^{k-2}(\text{ln}^{k-2}(r)^2^n)\]

For \(n=1\), the base case, it is trivial. The induction step goes as following:

\[f^{n+1}(r) = f(\text{exp}^{k-2}(\text{ln}^{k-2}(r)^2^n)) = \text{exp}^{k-2}(\text{ln}^{k-2}(\text{exp}^{k-2}(\text{ln}^{k-2}(r)^2^n))^2)=\text{exp}^{k-2}((\text{ln}^{k-2}(r)^2^n)^2) = \text{exp}^{k-2}(\text{ln}^{k-2}(r)^2^{n+1})\]

We know that \[\text{exp}^{k-2}(\text{ln}^{k-2}(r)^2^n)\] is k-exponential, because there are k+1 terms in total and the ln factor is to small to affect the topmost exponent. This solves the problem, q.e.d.