gpt pro 5.2 attempt on Math-5637

FORMAL RESTATEMENT

Definition (recursive rule). Define $f(1)=f(2)=1$. For $n\ge 3$, attempt to define \[ f(n) = f\big(n-f(n-1)\big)+f\big(n-f(n-2)\big), \] provided that the arguments $n-f(n-1)$ and $n-f(n-2)$ are positive integers for which $f(\cdot)$ has already been defined.

Well-definedness. Say the sequence is well-defined up to $N$ if for every $3\le n\le N$, both indices $n-f(n-1)$ and $n-f(n-2)$ lie in $\{1,2,\dots,n-1\}$.

Questions.

• $Q1$ Is $f(n)$ well-defined for all $n\in\mathbb{N}$?
• $Q2$ Does the range $\{f(n): n\in\mathbb{N}\}$ omit infinitely many positive integers?
• $Q3$ Describe the growth/behaviour of $f(n)$.

QUICK LITERATURE/CONTEXT CHECK

The extracted statement notes that this is asked by Hofstadter, that the sequence begins $1,1,2,3,3,4,\dots$ and is OEIS A005185, and that it is not known whether $f(n)$ is well-defined for all $n$.

ATTACK PLAN

Proof-track ideas (show well-defined for all $n$).

• Prove an inductive upper bound $f(n)\le n$ (or stronger) for all $n$; this would force $n+1-f(n)\ge 1$ and keep indices positive.
• Derive structural monotonicity or "almost monotonicity" properties to control growth.

Disproof-track ideas.

• Search computationally for the first $n$ such that either $n-f(n-1)\le 0$ or $n-f(n-2)\le 0$.
• For missing values, examine the set of attained values up to large $N$ and look for systematic gaps.

WORK

Fast reality check (computation)

Direct computation of the recursion (checking indices at each step) shows:

• The sequence is well-defined for all $n\le 1{,}000{,}000$.
• The first $50$ values (computed) are:

  1, 1, 2, 3, 3, 4, 5, 5, 6, 6,
  6, 8, 8, 8, 10, 9, 10, 11, 11, 12,
  12, 12, 12, 16, 14, 14, 16, 16, 16, 16,
  20, 17, 17, 20, 21, 19, 20, 22, 21, 22,
  23, 23, 24, 24, 24, 24, 24, 32, 24, 25
  

• For $N=200{,}000$, the maximum value attained is $\max_{n\le N} f(n)=108{,}656$. Among integers $1\le m\le 200$, the following $42$ values were not attained by $\{f(1),\dots,f(N)\}$:

  7, 13, 15, 18, 27, 29, 34, 36, 49, 51,
  59, 67, 70, 74, 81, 89, 95, 97, 98, 99,
  102, 103, 117, 126, 127, 131, 134, 141, 142, 145,
  150, 158, 163, 166, 181, 183, 189, 191, 195, 197, 198, 199
  

These computations do not prove infinite omission, but they show many gaps persist at least up to $N=200{,}000$.

Lemma 422.1 (Non-circularity and the only obstruction)

Lemma. Suppose $f(1),\dots,f(n-1)$ are defined and are positive integers. Then the only potential obstruction to defining $f(n)$ via \[ f(n)=f(n-f(n-1))+f(n-f(n-2)) \] is that one (or both) of the indices $n-f(n-1)$, $n-f(n-2)$ is $\le 0$. If both indices are $\ge 1$, then they automatically satisfy $\le n-1$ and the recursion uses only previously defined values.

Proof. Since $f(n-1)$ and $f(n-2)$ are positive integers, we have $f(n-1)\ge 1$ and $f(n-2)\ge 1$, hence \[ n-f(n-1)\le n-1,\qquad n-f(n-2)\le n-1. \] Therefore, if both indices are at least $1$, then they lie in the set $\{1,2,\dots,n-1\}$. By hypothesis, $f$ is already defined on $\{1,\dots,n-1\}$, so both terms on the right-hand side are defined, and the recursion determines $f(n)$ uniquely as a positive integer. Thus failure can only occur if an index is $\le 0$. $\square$

Lemma 422.2 (A uniform lower bound and characterisation of $f(n)=2$)

Lemma. For every $n\ge 3$ for which $f(n)$ is defined, one has $f(n)\ge 2$. Moreover, $f(n)=2$ if and only if both indices $n-f(n-1)$ and $n-f(n-2)$ lie in $\{1,2\}$.

Proof. Assume $f(n)$ is defined. Then both indices \[ a=n-f(n-1),\quad b=n-f(n-2) \] are positive integers, and \[ f(n)=f(a)+f(b). \] Since $f(1)=f(2)=1$ and (by definition of the recursion) all defined values of $f$ are positive integers, we have $f(a)\ge 1$ and $f(b)\ge 1$. Hence $f(n)=f(a)+f(b)\ge 2$.

Also, $f(n)=2$ holds if and only if $f(a)=f(b)=1$. Because the only indices with value $1$ among the initially specified values are $1$ and $2$ (i.e. $f(1)=f(2)=1$), this occurs exactly when $a,b\in\{1,2\}$. $\square$

VERIFICATION

• Lemma 422.1 explicitly isolates the issue: indices are always $<n$ (no circular dependence) once they are positive.
• Lemma 422.2 uses only positivity of the already-defined values and the initial conditions $f(1)=f(2)=1$.
• The computational checks explicitly verified the positivity of the indices at each step up to $n=10^6$.

FINAL

UNRESOLVED

1. (i) Strongest proved partial result here. Two rigorous structural lemmas:
   • Well-definedness fails only if an index becomes nonpositive; if indices stay positive then the recursion is non-circular and uniquely defines $f(n)$ from earlier values (Lemma 422.1).
   • For all defined $n\ge 3$, $f(n)\ge 2$, with a sharp characterisation of when equality holds (Lemma 422.2).
   Plus exact computation showing the recursion produces a well-defined sequence up to $n=1{,}000{,}000$, and that many integers (e.g. $7,13,15,\dots$) are not attained among $\{f(n):n\le 200{,}000\}$.
2. (ii) First gap (crisp). Either prove $f(n)$ is well-defined for all $n$ (e.g. by showing $f(n)\le n$ for all $n$, which would guarantee positivity of the indices), or else exhibit an explicit $n$ with $n-f(n-1)\le 0$ or $n-f(n-2)\le 0$.
3. (iii) Top 3 next moves.
   • Push computation further (e.g. to $n=10^7$) to look for any breakdown of the index-positivity condition, and to gather statistics on which integers are missed.
   • Attempt to prove an inductive upper bound of the form $f(n)\le c n$ with $c<1$ or at least $f(n)\le n$; any such bound would immediately imply global well-definedness.
   • Search for invariant or near-invariant quantities (e.g. relationships between $f(n)$ and $n$ modulo small integers, or bounds on successive differences) that could force persistent gaps in the range.
4. (iv) What a minimal counterexample would likely look like. A minimal counterexample to well-definedness would be the smallest $n$ such that $f(n-1)\ge n$ or $f(n-2)\ge n$, making an index nonpositive. Such a counterexample would require an unusually rapid growth spike in $f$ relative to $n$. A minimal counterexample to "misses infinitely many integers" would be evidence that beyond some point the range becomes cofinite; computationally, one would expect the set of missing values up to $\max_{k\le N} f(k)$ to shrink to zero density as $N\to\infty$.