My memory has cleared a little about Bell numbers, and I realise that what I had intended to mention earlier concerned their period modulo p, which can be shown to divide (p^p-1)/(p-1); I conjectured that it is exactly that for all p. I expect there are now factorisations available for a few more primes p, but haven't got around to checking on them ... Fred Lunnon On 3/1/06, Schroeppel, Richard <rschroe@sandia.gov> wrote:
This recurrence offers the intriguing possibility of calculating the Bell sequence with the Chinese Remainder Theorem. The product of primes < N is (very) roughly e^N. (Prime powers don't matter much here.) The asymptotic size of B(N) is rather larger, closer to N^N. So we'd need either additional congruence information or a small error estimate from the asymptotic formula (or more terms?) to continue. This isn't likely to be less work than simply using the recursion. [For the uninitiated, B's pop up in the power series for e^(e^x-1), and the recursion is just dotting the sequence-so-far with a row of binomial coefficients.]
Rich
-----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com on behalf of Fred lunnon Sent: Wed 3/1/2006 6:46 AM To: dasimov@earthlink.net; math-fun Subject: Re: [math-fun] Stirling Numbers? ... The Bell numbers B(n) have some nice properties modulo a fixed integer: for example, for prime p they satisfy B(n+p) - B(n+1) - B(n) = 0 (mod p); and for prime powers (E^p - E - 1)^k B(n) = 0 (mod p^k), in umbral notation where E denotes the shift operator E B(n) = B(n+1). ...