bibkey: karttunen2026a400168 authors: Antti Karttunen year: 2026 title: “OEIS A400168: arithmetic derivative primorial length minus the prime-power-divisor maximum” doi: null url: https://oeis.org/A400168 claim: “Question: Are there terms greater than 1?” strata_touched:

• D5/S3/Arith/OeisA400168PrimorialGap license: citation-only triage: anchor

The A400168 existence question

The sequence has offset one and NAME:

a(n) = A235224(A003415(n)) - A400164(n), where A235224 is the length of primorial base representation, A400164(n) = Max_{d|n, d is a prime power} A235224(n/d), and A003415 is the arithmetic derivative.

Its COMMENT asks:

Question: Are there terms greater than 1?

This is an existence question. The universal inequality used as the formal claim is its negative answer, not a conjecture asserted by Karttunen.

Write $P_0=1$ and let $P_k$ multiply the first $k$ primes. A002110 explicitly uses this indexing. A235224’s NAME sets $L(0)=0$ and, for positive $m$, takes the largest $k$ with $P_{k-1}\le m$. Equivalently, $L(m)$ is the least natural $k$ with $m<P_k$: strict increase gives $L(m)=j+1$ exactly when $P_j\le m<P_{j+1}$ for $m>0$. The least threshold exists without a finite cap, since $P_{m+1}\ge \mathrm{nthPrime}(m)\ge m+2$. The Lean definition uses Mathlib’s value-bounded primorial at the zero-based Nat.nth Nat.Prime k, not at the prime count $k$ itself.

A003415 sets $D(0)=D(1)=0$ and has the factorization formula $D(n)={\sum }_{p\mid n}{v}_p(n)(n/p)$ for positive $n$, equivalent to its prime values and product rule. The Lean definition uses the finite support of n.factorization; all its quotients are exact and the empty factorizations supply the zero and one cases.

A400164’s FORMULA maximizes $L(n/d)$ over every divisor in A246655, whose NAME specifies $p^e$ with $p$ prime and $e\ge 1$. Its COMMENT explicitly contrasts these with exponent-zero powers, excluding one. The finset supremum gives $M(1)=0$ from the empty filter and equals this maximum for all $n\ge 2$. The extension $M(0)=0$ is outside the offset-one domain.

Thus the original question is exactly $\exists n\ge 1,L(D(n))>M(n)+1$. Negating $\forall n\ge 1,L(D(n))\le M(n)+1$ has this same meaning. For natural lengths, either integer or truncated natural subtraction exceeds one precisely under the displayed strict inequality.

The exact input $n=2^{49}=562949953421312$ has $D(n)=13792273858822144$, $L(D(n))=15$ and $M(n)=13$, so its term is two. This answers the question positively without asserting that it is the first such input or that there are infinitely many.

Locator

• Original question: COMMENT %C A400168 Question: Are there terms greater than 1?
• Author and version: Antti Karttunen; A400168 revision 8, September 23, 2026, 08:49:51.
• Entry: https://oeis.org/A400168
• Official versioned export: https://github.com/oeis/oeisdata/blob/da8d37c6de95fccdc6caffee958a010bfd9172bc/seq/A400/A400168.seq
• Component definitions in that snapshot: A235224 NAME (revision 31), A002110 NAME and its zero convention, A003415 NAME/FORMULA and zero/one convention, A400164 FORMULA (revision 18), and A246655 NAME/COMMENT (revision 141).
• Component URLs: https://oeis.org/A235224 ; https://oeis.org/A002110 ; https://oeis.org/A003415 ; https://oeis.org/A400164 ; https://oeis.org/A246655
• Original preregistration: https://github.com/the-omega-institute/trureturing/issues/11883