arXiv subjects

Christian Hercher

Publications and source records attributed to Christian Hercher.

7 recordsLinked to original sources

arXiv2025

Castor Ministerialis

The famous problem of Busy Beavers can be stated as the question on how long a $n$-state Turing machine (using a 2-symbol alphabet or -- in a generalization -- a $m$-symbol alphabet) can run if it is started on the blank tape before it holds. Thus, not halting Turing machines are excluded. For up to four states the answer to this question is well-known. Recently, it could be verified that the widely assumed candidate for five states is in fact the champion. And there is progress in searching for good candidates with six or more states. We investigate a variant of this problem: Additionally to the requirement that the Turing machines have to start from the blank tape we only consider such Turing machines that hold on the blank tape, too. For this variant we give definitive answers on how long such a Turing machine with up to five states can run, analyze the behavior of a six-states candidate and give some findings on the generalization of Turing-machines with $m$-symbol alphabet.

cs.FL↗

arXiv2025

On one of Erdős' Problems -- An Efficient Search for Benelux Pairs

Erdős asked for positive integers $m 2^{40}$. For the analogous problem of integers $m<n$ with $m$ and $n+1$ having the same set of prime factors and $m+1$ and $n$having the same set of prime factors, the situation is very similar: An infinite series and one exceptional solution with $n\leq 2^{22}+2^{12}\approx 4.2\cdot 10^6$ were known. We prove that there are no other exceptional solutions with $n<1.4\cdot 10^{12}$.

math.NT↗

arXiv2025

On Positive Integers $n$ with $ϕ(n)=\frac{2}{3} \cdot (n+1)$

While solving a special case of a question of Erdős and Graham Steinerberger asks for all integers $n$ with $ϕ(n)=\frac{2}{3} \cdot (n+1)$. He discovered the solutions $n\in\{5, 5 \cdot 7, 5\cdot 7\cdot 37, 5\cdot 7\cdot 37\cdot 1297\}$ and found that any additional solution must be greater than $10^{10}$. He conjectured that there are no such additional solutions to this problem. We analyze this problem and prove: *) Every solution $n$ must be square-free. *) If $p$ and $q$ are prime factors of a solution $n$ then $p\nmid (q-1)$. *) Any solution additional to the set given by Steinerberger has to have at least 7 prime factors. *) For any additional solution it holds $n\geq 10^{14}$.

math.NT↗

arXiv2024

Triangular Numbers With a Single Repeated Digit

The question of which triangular numbers have a decimal representation containing a single repeated digit seamed to be settled since at least the 1970s: Ballew and Weger provided a complete list and a proof that these are the only numbers of this kind. This assertion is referenced by other authors in the field. However, their proof is flawed. We present a new and elementary proof of the statement, which corrects the error.

math.NT↗

arXiv2024

On the Sum of Squarefree Integers and a Power of Two

Erdos conjectured that every odd number greater than one can be expressed as the sum of a squarefree number and a power of two. Subsequently, Odlyzko and McCranie provided numerical verification of this conjecture up to $10^7$ and $1.4\cdot 10^9$. In this paper, we extend the verification to all odd integers up to $2^{50}$, thereby improving the previous bound by a factor of more than $8\cdot 10^5$. Our approach employs a highly parallelized algorithm implemented on a GPU, which significantly accelerates the process. We provide details of the algorithm and present novel heuristic computations and numerical findings, including the smallest odd numbers $<2^{50}$ that require a higher power of two as all smaller ones in their representation.

math.NT↗

arXiv2023

Efficient Calculation the Number of Partitions of the Set $\{1, 2, \ldots, 3n\}$ into Subsets $\{x, y, z\}$ Satisfying $x+y=z$

Consider the set $\{1,2,\ldots,3n\}$. We are interested in the number of partitions of this set into subsets of three elements each, where the sum of two of them equals the third. We give some criteria such a partition has to fulfill, which can be used for efficient pruning in the search for these partitions. In particular, we enumerate all such partitions for $n=16$ and $n=17$ adding new terms to the series A108235 in the Online Encyclopedia of Integer Sequences.

math.CO↗

arXiv2023

There are no Collatz-m-Cycles with $m\leq 91$

The Collatz conjecture (or ``Syracuse problem'') considers recursively-defined sequences of positive integers where $n$ is succeeded by $\tfrac{n}{2}$, if $n$ is even, or $\tfrac{3n+1}{2}$, if $n$ is odd. The conjecture states that for all starting values $n$ the sequence eventually reaches the trivial cycle $1, 2, 1, 2, \ldots$ . We are interested in the existence of nontrivial cycles. Let $m$ be the number of local minima in such a nontrivial cycle. Simons and de Weger proved that $m \geq 76$. With newer bounds on the range of starting values for which the Collatz conjecture has been checked, one gets $m \geq 83$. In this paper, we prove $m \geq 92$. The last part of this paper considers what must be proven in order to raise the number of odd members a nontrivial cycle has to have to the next bound -- that is, to at least $K\geq1.375\cdot 10^{11}$. We prove that it suffices to show that, for every integer smaller than or equal to $1536\cdot2^{60}=3\cdot2^{69}$, the respective Collatz sequence enters the trivial cycle. This reduces the range of numbers to be checked by nearly $60$\%.

math.NT↗