Let d(n) be the Conway-type integer sequence. For K≥ -1, let I_K=n:2^K+1≤ n≤ 2^K+2, with initial domain I_-1=1,2. For each n, let n_1 and n_2 be the two parent spots occurring in the recurrence for d(n). Parent-spot generation conjecture. For every n∈ I_K with K≥ 1, both parent spots satisfy n_1,n_2∈ I_K-1. This describes the generation structure underlying the recurrence for d(n). The source reports that the claim has been numerically verified to high order of K, but gives no proof or further resolution.

The parent-spot generation conjecture for the Conway-type sequence d(n)

Less than 1 year old · traced to 2026

Let $d(n)$ be the Conway-type integer sequence. For $K\geq -1$ , let

I_K=\left\{n:2^{K+1}\leq n\leq 2^{K+2}\right\},

with initial domain $I_{-1}=\{1,2\}$ . For each $n$ , let $n_1$ and $n_2$ be the two parent spots occurring in the recurrence for $d(n)$ . Parent-spot generation conjecture. For every $n\in I_K$ with $K\geq 1$ , both parent spots satisfy

n_1,n_2\in I_{K-1}.

This describes the generation structure underlying the recurrence for $d(n)$ . The source reports that the claim has been numerically verified to high order of $K$ , but gives no proof or further resolution.

References

Primary source

Klaus Pinn, “Study of Meta-Fibonacci Integer Sequences by Continuous Self-Referential Functional Equations”, arXiv:2603.17509 (2026).

integer-sequences meta-fibonacci-sequences

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

The conjecture has been numerically supported, but a posted calculation claims a boundary counterexample that would disprove it; this has not been independently checked.

The conjecture places both recursive parent positions of every term in generation $K$ inside the preceding generation interval. Klaus Pinn’s 2026 study reports high-order numerical verification but gives no proof or resolution.

Posted attempt

A reader-written calculation claims a complete counterexample at $K=3$ and $n=16$ : one parent is $8$ , while the other is $7$ , outside the closed interval $I_2=[8,16]$ . If correct, this disproves the conjecture as stated; the calculation has not been independently verified.

March 2026 numerical verification

Pinn reports that the parent-spot generation pattern was numerically verified to high order of $K$ , but records no proof and does not settle the boundary issue raised by the later calculation.

Current status (as of August 2026): Numerical evidence supports the pattern, but the conjecture is now challenged by an unverified boundary counterexample, so neither the conjecture nor its disproof is settled.

Sources

arxiv.org

CounterexampleThis solution needs a summarySee full solutionHide full solution

Let

D(1)=D(2)=1,\qquad D(n)=D(D(n-1))+D(n-1-D(n-2)),

and set $d(n)=2D(n)-n$ . The two parent spots are

n_1=D(n-1)=\frac{n-1+d(n-1)}2,\qquad n_2=n-1-D(n-2)=\frac{n-d(n-2)}2.

The generation intervals in the conjecture are explicitly closed and overlapping:

I_K=\{n:2^{K+1}\le n\le2^{K+2}\}.

Direct iteration of the defining recurrence gives

(D(1),\ldots,D(16)) =(1,1,2,2,2,3,4,4,4,4,5,6,7,8,8,8).

Now take $K=3$ and $n=16$ . Then

n\in I_3=[16,32],\qquad I_2=[8,16],

but

n_1=D(15)=8\in I_2,\qquad n_2=15-D(14)=7\notin I_2.

Equivalently, $d(14)=2$ , and hence $n_2=(16-2)/2=7$ .

The same pair of parent spots $(8,7)$ appears in Table 1 of the source, which explicitly notes that boundary indices belong to two consecutive generation intervals. Therefore the asserted universal inclusion is false for the stated closed intervals. A version with half-open intervals or an exception at lower endpoints would be a different conjecture.