A341617 Repair factors for Stirling numbers of the second kind.
1, 1, 2, 6, 12, 60, 30, 210, 840, 2520, 1260, 13860, 13860, 180180, 90090, 30030, 240240, 4084080, 6126120, 116396280, 58198140, 58198140
Offset: 1
Examples
The statement that a(3) = 2 means that the sequence of Stirling numbers S_3 = (1, 6, 25, 90, ...) (that is, the sequence A000392 with an offset of 3) does not have the property of counting periodic points for some map, but does have this property after multiplication by 2 (which gives A028243 with an offset of 2), and 2 is the smallest integer with this property. This specific value is immediately known to be exact, because a(3) divides (3-1)! = 2.
Links
- P. Miska and T. Ward, Stirling numbers and periodic points, arXiv:2102.07561 [math.NT], 2021.
Formula
It is known that a(k) divides (k-1)!.
There is an explicit formula for a(k), but it involves an in principle infinite calculation as follows: compute the set of rational numbers {(1/n) Sum_{d|n} mu(n/d)*S(d+k-1, k): n>=1}, and then define a(k) to be the least common multiple of the denominators of that (possibly infinite) set of rational numbers. Here mu denotes the classical Möbius function, and the sum is taken over the divisors d of n. Strictly speaking, any calculation only gives candidate values which are initially only known to be a factor of the real value, which in turn is a factor of (k-1)!. For small values of k listed above ad hoc arguments are needed to check the candidate values evaluated as the least common multiple of the denominators of the first 3000 terms.
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