This seems like it might appeal to some folks here. Rich rcs@cs.arizona.edu -----Original Message----- From: Number Theory List on behalf of Don Davis Sent: Fri 7/8/2005 8:38 PM To: NMBRTHRY@LISTSERV.NODAK.EDU Subject: divisibility conjectures Some conjectures about divisibility by 2 have arisen in my work in algebraic topology. The first one involves Stirling numbers of the second kind. It would give a nice form for homotopy groups of some important topological spaces. We write nu(-) for the exponent of 2 in an integer. Conj 1: If j \ge n and L is sufficiently large (e.g. L>3n/2), then nu(j! S(2^L + n -1, j)) \ge n-1 + nu([n/2]!). It is not difficult to remove the 2^L and the n-1 on the RHS, obtaining the following conjecture, which implies Conj 1. Conj 2: If j>m, then nu(sum ((j choose 2k) k^m)) \ge nu([(m+1)/2]!). Instead of dealing with expressions k^m, one might prefer to work entirely with binomial coefficients. We have the following conjecture, which implies Conj 2. Our sums are taken over all sensible values of k. Conj 3: nu( sum ((j choose (4k+2)) (k choose i))) \ge nu([j/2]!) - i - nu(i!). This can be interpreted as a coefficient in a power series, of which it appears that not only the coefficient in question is highly divisible, but also all subsequent coefficients. We obtain the following conjecture, which implies Conj 3. Conj 4: nu( coeff( (1+x)^j / (1-x^4)^(i+1), x^k) ) \ge nu([j/2]!) - i - nu(i!) for k > j-4i-4. It is quite dramatic that the coefficient of x^(j-4i-4) will be odd, but then apparently all the rest are highly divisible. It is enough to prove this when j = 4a, and probably when i+1 is a 2-power. Experimenting with related power series, by removing the 4's, we are led to the following conjecture, which is closely related to Conj 4. My feeling is that this last conjecture is one which others might have seen, and that a proof of it would probably lead to a proof of Conj 4. Conj 5: If 2^t > 2d, then nu( coeff( ((1+x)/(1-x))^(2^t) (1+x)^(2d), x^k) ) \ge t + d + nu(d!) + 1 if k > 2d. Just to clarify...the power series in Conj 5 equals (1+x)^(2^t + 2d) / (1-x)^(2^t). If anyone on this listserve can point to any literature related to these conjectures, or offer suggestions toward a proof, I would be most appreciative. If you can prove one, we can discuss an appropriate way to publish it. Don Davis dmd1@lehigh.edu http://www.lehigh.edu/~dmd1