Prime support and growth of Stirling repair factors
Henry Zweiman
September 8, 2026
Abstract
For fixed , let be the least positive integer such that counts the periodic points of a map, where is a Stirling number of the second kind. We prove
Thus the prime divisors of are exactly the primes below , and the previous factorial bound can be replaced by a least common multiple. We also prove whenever , and for every prime and . A formula for prime-power values in terms of the base- digits of also yields . These results establish Conjectures 1, 2, 3, and 4(d) of Miska and Ward. The upper bound follows from a Frobenius congruence for an integral recurrence; the prime-support obstruction comes from the repeated poles of its generating function modulo .
1. Introduction
An integer sequence satisfies the Dold condition if
For a nonnegative integer sequence, realizability as the numbers of fixed points of the iterates of a map is equivalent to (1) together with nonnegativity of the sums in (1). Indeed, their quotients by are precisely the required numbers of orbits of length . A disjoint union of those finite cycles supplies a realizing map.
Miska and Ward introduced the least positive multiplier making a sequence realizable and studied it for Stirling sequences [3]. Throughout this paper,
They proved that is finite and divides [3, Theorem 2.2]. Their Conjectures 1 and 2 predict, respectively, that every prime below divides , and that primes in have valuation exactly one. Their Conjecture 4(d) predicts . Their Conjecture 3 predicts that tends to infinity. We prove all four assertions, improve the factorial upper bound, and determine the logarithmic growth of that ratio.
Theorem 1.
For every integer ,
where . More precisely, for every prime ,
In addition:
if , then ;
if with , then .
Theorem 2.
Let , where the sum is over primes. As , with ,
Consequently,
The estimate (3) is elementary. Only the passage to (4) uses the prime number theorem.
The upper bound is sharp at the prime whenever . The argument for prime support applies to the entire sequence and does not require locating an individual prime-power witness. In particular, our result does not establish the proposed finite cutoff for computing described in [1, Section 4.5].
2. Local congruences and the integral recurrence
For a sequence write . The following standard local form of the Dold condition will be useful; we include the proof to fix the indexing.
Lemma 3.
An integer sequence satisfies the Dold condition if and only if
for every prime , every , and every with .
Proof.
Möbius inversion gives . Thus
which proves one implication. Conversely,
The congruences imply . Applying this at each prime factor of an arbitrary index gives (1).
Lemma 4.
For every , the sequence satisfies for all . Consequently, is its least positive Dold multiplier.
Proof.
For the sequence is constant equal to one, so the assertion is immediate. For , the Stirling recurrence gives
As , for we have
Therefore . Positive multiplication preserves this sign condition.
Fix , and suppress the superscript on . Define
The usual generating identity for Stirling numbers is
It follows directly, for example, by applying the Stirling recurrence to : this series is the corresponding series for , divided by , and the initial series for is .
Lemma 5.
Set . There are an integer matrix , a vector , and an integer linear functional such that
For this uses the coefficient of , which is zero since , and the fact that . For it uses the coefficient of . Take the companion transition matrix for the states , take the initial state as , and let extract its first coordinate. The companion identity gives the result.
The integral initialization at zero is useful: the matrix powers represent itself, without a shift in the exponent.
3. The least-common-multiple upper bound
Lemma 6.
Let be commuting integer matrices. If , where is prime and , then
Proof.
Write ; then commutes with . In the binomial expansion of , the terms of degree in , for , are divisible by , hence by . The last term is divisible by , and . Iteration proves the assertion, including .
Every residue class modulo occurs among at most times. Over this implies , so modulo . Since , raising to the power gives zero. The characteristic- binomial identity for commuting matrices gives modulo .
The case is immediate. For , fix a prime and use the notation of Lemma 7. For , Lemma 6 gives
Taking arbitrary positive powers preserves the congruence. Applying therefore gives
Multiplication by supplies the required local modulus . For the same conclusion is automatic after multiplication by .
If , then and . If , then
Indeed, for the integer on the right, , and hence . The product of these local multipliers is
Lemmas 3 and 4 show that this is a realizability multiplier. The least multiplier divides every multiplier: it is the least common multiple of the denominators of the rational numbers in lowest terms. Consequently .
Let , and suppose that . The local congruences for , reduced modulo , imply
Here one writes with and applies the local congruence at level , cancelling the nonzero scalar modulo . Iteration gives for every .
Choose as in Lemma 7 and set modulo . Then , whence for every . The matrix representation and (8) imply
Thus the generating series modulo has period , and is a polynomial. By (6), this forces
There is no numerator cancellation because the numerator in (6) is the constant one. But means that the factors with and both occur, so modulo . The polynomial is squarefree: its derivative in characteristic is , which has no common root with it. This is a contradiction, also when . Hence for every .
5. Prime-power values and exact valuations
The next formula extends the first-level congruence of Miska and Ward [3, proof of Lemma 2.5].
Lemma 8.
Let be prime, , and . Write
Then
where the empty product is one.
Proof.
Write , with , and put . Over we have
If , every coefficient of at a nonnegative degree divisible by equals one modulo . Indeed, the coefficient is , and inclusion–exclusion gives
Here is invertible modulo and Fermat’s congruence applies to each . Thus, when and , the coefficient of in is
When and , this coefficient is one, which agrees with the final binomial expression. If , only the degree-zero coefficient of contributes, and the answer is . Putting combines both cases into
The base- digits of are , and those of are one in positions and zero thereafter. If all are at most , adding and produces no carry. Lucas’s coefficient formula therefore evaluates (10) as . If one of these digits is , take the first such digit. There was no earlier carry, and the corresponding digit of is zero. Its Lucas factor is , and the proposed product is zero as well. For completeness, the coefficient formula used here follows by comparing coefficients in
where are the base- digits of .
Corollary 9.
Let , put , and use the digits in Lemma 8. If , , and for every , then . In particular, if for every , then
Under the stated conditions this is nonzero. The local Dold congruence at level forces . At , the highest digit is nonzero, and the upper bound in Theorem 1 gives equality.
Remark 10.
Lemma 8 also supplies an explicit witness for full prime support. For , let be the first position with . Then , with . This proves the prime-support assertion by a second method and locates one necessary prime-power factor below . It does not locate the maximum possible valuation of the repair factor.
When , we have , so the digit restrictions in Corollary 9 are empty and . When , the digits of are zero below position and one at position . The same corollary gives . In particular, for the corollary also gives at both and , for every .
For the error estimate, there are terms with , and each is at most , by the elementary inequality .
For the lower bound, consider primes
For each such prime, . Corollary 9 gives unless the digit in is . In that exceptional case, satisfies
For a fixed , at most one positive integer can satisfy (11). Indeed, the lower endpoint for exceeds the upper endpoint for , since
The intervals increase with , so they are pairwise disjoint. Consequently, at most primes in the specified range are exceptional.
Every nonexceptional prime contributes at least to . It follows that
This proves (3). The classical prime number theorem [2], together with , proves (4). In particular, .
7. Remaining questions
The exact value of for general is still not determined by these arguments. In particular, the finite-cutoff assertion discussed in [1, Section 4.5] would compute from the denominators of only for indices at most the largest prime power strictly below . Our upper bound controls all prime-power valuations, but does not show that their maxima occur within that finite range.
The results above leave open parts (a)–(c) of Conjecture 4 in [3]. In particular, the assertion for requires a sharper upper bound than the one proved here. Distinguishing those special cancellations from the general Frobenius bound appears necessary for an exact formula.
References
[1] J. Byszewski, G. Graff, and T. Ward, Dold sequences, periodic points, and dynamics, Bull. London Math. Soc. 53 (2021), 1263–1298. doi:10.1112/blms.12531.
[2] J. Hadamard, Sur la distribution des zéros de la fonction et ses conséquences arithmétiques, Bull. Soc. Math. France 24 (1896), 199–220. doi:10.24033/bsmf.545.