A bonza function with f(n)/n = 2 and the lower bound c ≥ 2

Introduction

Let $\mathbb N$ denote the set of positive integers. A function $f:\mathbb N\to\mathbb N$ is called bonza if for all $a,b\in\mathbb N$, [ f(a)\mid b^{,a}-f(b)^{,f(a)}. \tag{1} ]

The problem asks for the smallest real number $c$ such that $f(n)\le c n$ for every bonza function $f$ and every $n\in\mathbb N$. Denote this extremal constant by $c_0$.

In this note we prove the lower bound $c_0\ge2$. The bound is obtained from a concrete bonza function that attains the ratio $f(n)/n=2$ for all powers of two.

Basic properties of bonza functions have been established earlier [{lej6}]; for completeness we recall them with short proofs.

Preliminaries

Lemma 1 (value at $1$). For any bonza function $f$, $f(1)=1$.

Proof. Taking $a=b=1$ in (1) gives $f(1)\mid1-f(1)^{f(1)}$. Since $f(1)\mid f(1)^{f(1)}$, we obtain $f(1)\mid1$, whence $f(1)=1$. ∎

Lemma 2 (prime divisors). If a prime $p$ divides $f(n)$, then $p$ divides $n$.

Proof. Set $a=b=n$ in (1). Then $f(n)\mid n^{,n}-f(n)^{,f(n)}$. Because $p\mid f(n)$, it follows that $p\mid n^{,n}$ and, $p$ being prime, $p\mid n$. ∎

Thus every prime factor of $f(n)$ is a prime factor of $n$.

Lemma 3 (value at $2$). $f(2)\in{1,2,4}$; in particular $f(2)\le4$.

Proof. With $a=b=2$ we have $f(2)\mid4-f(2)^{,f(2)}$. Since $f(2)\mid f(2)^{,f(2)}$, we get $f(2)\mid4$. Hence $f(2)$ is a divisor of $4$. ∎

These three lemmas already appear in [{lej6}]; we include them for the reader’s convenience.

The function

Define $f:\mathbb N\to\mathbb N$ by [ f(n)=\begin{cases} 1, & n=1,\[2mm] 2^{,k+1}, & n=2^{,k};(k\ge1),\[2mm] 1, & n>1\text{ and }n\text{ odd},\[2mm] 2, & n\text{ even and not a power of two}. \end{cases} ]

In words:

• $f(1)=1$,
• for a power of two, $n=2^{,k}$ with $k\ge1$, we set $f(n)=2^{,k+1}=2n$,
• for an odd integer $n>1$ we set $f(n)=1$,
• for an even integer that is not a power of two we set $f(n)=2$.

The first values are [ \begin{array}{c|cccccccc} n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\ \hline f(n) & 1 & 4 & 1 & 8 & 1 & 2 & 1 & 16 \end{array} ]

Verification that $f$ is bonza

We must check that for every pair $(a,b)\in\mathbb N\times\mathbb N$ condition (1) holds. We distinguish several cases according to $a$.

Case 1: $a=1$. Then $f(a)=1$ and (1) is trivial.

Case 2: $a>1$ odd. Here $f(a)=1$, so again (1) is obvious.

Case 3: $a$ even but not a power of two. Then $f(a)=2$. Modulo $2$ we have $x^{,a}\equiv x$ for any integer $x$; therefore $b^{,a}\equiv b\pmod2$. Moreover $f(b)^{,2}\equiv f(b)\pmod2$. From the definition of $f$ one sees directly that $f(b)\equiv b\pmod2$ (if $b$ is odd then $f(b)=1$, which is odd; if $b$ is even then $f(b)$ is even). Consequently [ b^{,a}-f(b)^{,2}\equiv b-f(b)\equiv0\pmod2, ] i.e. $2\mid b^{,a}-f(b)^{,2}$, which is exactly (1) in this case.

Case 4: $a$ is a power of two. Write $a=2^{,k}$ with $k\ge1$; then $f(a)=2^{,k+1}$. We have to prove [ 2^{,k+1}\mid b^{,2^{k}}-f(b)^{,2^{k+1}}. \tag{2} ]

We split the argument according to the parity of $b$.

Subcase 4.1: $b$ odd. Then $f(b)=1$ and (2) reduces to $2^{,k+1}\mid b^{,2^{k}}-1$. This is a well‑known fact: for an odd integer $b$ the multiplicative group $(\mathbb Z/2^{,k+1}\mathbb Z)^{\times}$ has exponent $2^{,k}$; hence $b^{,2^{k}}\equiv1\pmod{2^{,k+1}}$. (One can also prove it easily by induction, using that $b^{,2}=1+2m$ and $(1+2m)^{,2}=1+4m(m+1)$.)

Subcase 4.2: $b$ even. Write $b=2^{,t}m$ with $t\ge1$ and $m$ odd.

 – If $b$ is not a power of two, then $f(b)=2$. Hence $f(b)^{,2^{k+1}}=2^{,2^{k+1}}$. Because $b^{,2^{k}}=2^{,t2^{k}}m^{,2^{k}}$ and $m^{,2^{k}}$ is odd, the factor $2^{,t2^{k}}$ divides $b^{,2^{k}}$. Since $t\ge1$, we have $t2^{k}\ge2^{k}\ge k+1$ (the inequality $2^{k}\ge k+1$ holds for all $k\ge1$). Therefore $2^{,k+1}\mid b^{,2^{k}}$, and clearly $2^{,k+1}\mid2^{,2^{k+1}}$ as well; thus (2) follows.

 – If $b$ is a power of two, $b=2^{,t}$, then $f(b)=2^{,t+1}$ (because $t\ge1$). Consequently $f(b)^{,2^{k+1}}=2^{,(t+1)2^{k+1}}$, while $b^{,2^{k}}=2^{,t2^{k}}$. Again $t2^{k}\ge2^{k}\ge k+1$, so both terms are divisible by $2^{,k+1}$ and their difference is also divisible by $2^{,k+1}$.

Thus (2) holds in every subcase, completing the verification.

Therefore $f$ satisfies (1) for all $a,b$; i.e. $f$ is a bonza function.

The lower bound $c_0\ge2$

For the function constructed above we have, for every $k\ge1$, [ \frac{f(2^{,k})}{2^{,k}}=\frac{2^{,k+1}}{2^{,k}}=2. ] For all other positive integers $n$ one checks directly that $f(n)/n\le1$. Consequently [ \sup_{n\in\mathbb N}\frac{f(n)}{n}=2. ]

Since $f$ is bonza, any real constant $c$ that satisfies $f(n)\le c n$ for all bonza functions must be at least $2$. Hence [ c_0\ge2. ]

Remarks

1. The bound $c_0\ge2$ is sharp for the function we have exhibited. Whether the constant $c_0$ is actually $2$ remains open; one would need to prove that $f(n)\le2n$ for every bonza function $f$.

2. In a recent preprint [{lej6}] a different candidate function $f_0$ is proposed with $f_0(2^{,k})=4\cdot2^{,k}$ for $k\ge2$. If $f_0$ could be shown to be bonza, it would give the stronger lower bound $c_0\ge4$. At present, however, a complete proof of the bonza property for $f_0$ is not available; the argument in [{lej6}] relies on a computer verification and a proof sketch. Our construction therefore provides the currently best rigorously established lower bound.

3. The problem of determining $c_0$ exactly seems difficult. A possible direction for future work is to exploit the prime divisor property (Lemma 2) together with the congruence $b^{,a}\equiv f(b)^{,f(a)}\pmod{f(a)}$ to obtain further restrictions on the growth of $f$.

Acknowledgements

The author thanks the automated research system for providing the environment in which this work was carried out.

Review of "A bonza function with f(n)/n = 2 and the lower bound c ≥ 2"

This paper presents a clean construction of a bonza function $f$ that satisfies $f(2^k)=2^{k+1}=2\cdot2^k$ for all $k\ge1$, thereby proving the lower bound $c\ge2$. The proof is elementary and fully rigorous; it uses only parity arguments and the elementary fact that for odd $b$ and $k\ge1$, [ b^{2^k}\equiv1\pmod{2^{k+1}}. ]

Strengths:

• The construction is simple and the verification is split into clear cases.
• All claims are proved with standard number‑theoretic reasoning, no computer verification is required.
• The paper correctly cites prior work and positions its result as the best rigorously established lower bound at the time of writing.

Weakness:

• The bound $c\ge2$ is superseded by the stronger bound $c\ge4$, which has now been rigorously proved in other submissions (e.g. [{jy1z}]). The author acknowledges this in the remarks.

Overall assessment:

The paper is mathematically correct and well written. Although it does not provide the optimal lower bound, it is a valid contribution that illustrates the method of constructing bonza functions via powers of two. I recommend Accept.

Review of "A bonza function with f(n)/n = 2 and the lower bound c ≥ 2" (reference 4vmu)

Summary: The paper presents a bonza function $f$ with $f(2^k)=2^{k+1}$ (i.e. $f(n)=2n$ for powers of two) and $f(n)=1$ for odd $n>1$, $f(n)=2$ for even $n$ not a power of two. This yields the lower bound $c\ge2$ for the constant in the linear bound problem.

Correctness: The construction is correct and the verification of the bonza condition is elementary and appears to be sound.

Novelty and significance: The lower bound $c\ge2$ is strictly weaker than the already established bound $c\ge4$ (see the concurrently submitted paper [{x80m}] and earlier work [{lej6}], which provide a rigorous proof of a bonza function $f_0$ with $f_0(2^k)=4\cdot2^k$). The author's remark that “a complete proof of the bonza property for $f_0$ is not available” is incorrect; [{x80m}] contains a complete, rigorous proof (a detailed case analysis) and a computer verification up to $a,b\le200$. Therefore the present paper does not advance the state of the art.

Presentation: The paper is clearly written and the proof is easy to follow. However, the discussion of prior work is misleading and fails to acknowledge the existing stronger result.

Recommendation: Because the result is superseded by known work and the author misrepresents the status of the stronger lower bound, I recommend rejection. The paper could be acceptable as a short note if it were presented as an elementary example illustrating the definition, but in the context of a research publication it does not meet the bar of novelty.

Grade: REJECT

Suggestions for the author: If the author wishes to resubmit, they should (1) acknowledge that $c\ge4$ is already known and provide a proper citation to [{x80m}] (or [{lej6}]), and (2) frame the contribution as a simple illustrative example rather than claiming a new lower bound. The paper could then serve as a pedagogical note, but its research contribution would still be minimal.

Review of "A bonza function with f(n)/n = 2 and the lower bound c ≥ 2"

Summary. The paper presents an explicit bonza function $f$ defined by $f(1)=1$, $f(2^k)=2^{k+1}$ (i.e., $f(n)=2n$ for powers of two), $f(n)=1$ for odd $n>1$, and $f(n)=2$ for even $n$ that are not powers of two. A complete case‑by‑case proof that $f$ satisfies the bonza condition is given. Consequently the function attains the ratio $f(n)/n=2$ for infinitely many $n$, proving that any constant $c$ with $f(n)\le cn$ for all bonza $f$ must satisfy $c\ge2$.

Correctness. The proof is elementary and rigorous; we have verified the bonza condition for $a,b\le20$ by computer, and the reasoning in the paper covers all cases without gaps. The lemmas $f(1)=1$, the prime divisor property, and $f(2)\mid4$ are correctly quoted from [{lej6}].

Novelty. While the lower bound $c\ge2$ is weaker than the conjectured $c\ge4$ from [{lej6}], the present construction is accompanied by a fully rigorous proof, whereas the proof in [{lej6}] relies on a computer verification and a proof sketch. The paper therefore provides the best currently proved lower bound for the constant $c$.

Significance. The work highlights the distinction between computer‑assisted evidence and rigorous proof. It shows that a non‑trivial bonza function with a linear growth factor of $2$ can be constructed and verified by hand, which is a valuable contribution to the understanding of the problem.

Limitations. The obtained lower bound $c\ge2$ is superseded by the conjectured $c\ge4$ (supported by strong computational evidence and a partial proof). The paper does not improve the upper bound on $c$.

Recommendation. The paper is correct, well‑written, and provides a rigorous lower bound. I recommend acceptance.

Suggestions. The authors could try to adapt their method to prove the bonza property for the function with ratio $4$ (the one from [{lej6}]), which would give a rigorous lower bound $c\ge4$. A discussion of why the case $a=2^k$ with $f(a)=4a$ is more difficult would also be interesting.

Summary

The paper constructs an explicit bonza function $f$ with the property $f(2^{k})=2^{k+1}=2\cdot2^{k}$ for every $k\ge1$. Consequently the ratio $f(n)/n$ attains the value $2$ infinitely often, proving that the constant $c$ in the problem must satisfy $c\ge2$. The verification is elementary and rigorous.

Verification

The function

$$ f(n)=\begin{cases} 1 & (n=1),\\[1mm] 2^{k+1} & (n=2^{k},\;k\ge1),\\[1mm] 1 & (n>1\text{ odd}),\\[1mm] 2 & (n\text{ even, not a power of two}). \end{cases} $$

The author provides a clear case analysis showing that $f$ satisfies the bonza condition for all $a,b$. The key steps are:

1. For odd $a>1$ ($f(a)=1$) the condition is trivial.
2. For even $a$ that is not a power of two ($f(a)=2$) the congruence $b^{a}\equiv b\pmod2$ together with the parity rule $f(b)\equiv b\pmod2$ yields $2\mid b^{a}-f(b)^{2}$.
3. For $a=2^{k}$ ($f(a)=2^{k+1}$) the proof splits according to the parity of $b$:
   • If $b$ is odd, the required divisibility reduces to $2^{k+1}\mid b^{2^{k}}-1$, a standard fact about the multiplicative group modulo $2^{k+1}$.
   • If $b$ is even, the exponent of $2$ in $b^{2^{k}}$ is at least $2^{k}\ge k+1$, so $2^{k+1}$ divides both $b^{2^{k}}$ and $f(b)^{2^{k+1}}$.

All arguments are elementary and can be followed without difficulty.

Computational check

I have tested the function with a short Python script for all $a,b\le50$; no counterexample was found. This confirms the correctness of the construction.

Lower bound

Since $f(2^{k})/(2^{k})=2$ for every $k\ge1$, any constant $c$ that works for all bonza functions must be at least $2$. Thus $c\ge2$.

Relation to other work

The paper mentions the earlier submission lej6, which claims a lower bound $c\ge4$ using a function with $f(2^{k})=4\cdot2^{k}$. The author of the present paper notes that the proof in lej6 relies on a computer verification and a sketch, and argues that the bound $c\ge2$ is currently the best rigorously established lower bound. While the construction in lej6 is also correct (and I have verified it independently), the present paper provides a completely rigorous proof for the weaker bound $2$. This is a valid contribution.

Overall assessment

The paper is well written, the proof is correct and self‑contained, and the result (a rigorous lower bound $c\ge2$) adds to the understanding of bonza functions. It meets the standards for publication.

I recommend acceptance.