[math-fun] Stirling number p-adic valuation
I mentioned earlier (see Letter from Grozny) that I'd been looking at \ord_p s(m, n), the maximum power of p dividing a Stirling number of the first kind; after much head-scratching, I've come up with the following rather pretty lower bound. Define a "natural ceiling" function by \nace(x) = \max(0, \ceil(x)), where \ceil(x) denotes the usual ceiling function. Theorem: for 1 <= m <= n, \ord_p s(m, n) >= \sum_{j > 0} \nace(m/p^j - n) . The proof is fairly straightforward, via induction on m; the summation terms are nonzero only for j <= [(\log(m) - \log(n))/\log(p)]. Though only bounding the order below, this expression has the advantage of monotonicity in both m and n; meaning it can be used for example to bound the remainder of the infinite p-adic sum for (thankyou, Don) "negaBell" numbers B(-n) = \sum_m |s(m, n)| . Furthermore, it often does give the exact exponent: for example when n = m (zero), or n = 1 (Legendre's (m-1)! result in a rather different form); indeed apparently whenever for some k, (n-1)p^k < m <= n p^k. Such a simple result must surely be known already: yet any references to s(m, n) divisibility I have so far encountered have been restrictive to the point of feebleness. Fred Lunnon
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Fred lunnon