Timeline for A binomial sum is divisible by p^2
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| Jan 11, 2010 at 14:01 | comment | added | darij grinberg | ... are easily seen to be divisible by u, since $\binom{2k}{k}$ is divisible by u for every k between p and 2p-2). So the numerator must be divisible by all primes u between p and 2p-1, including p twice (according to the original question). With so many little prime factors, there is not much place for big prime factors - but this is just a heuristic argument, and I wouldn't be surprised if for really large p, the big factors would "win" (i. e., their distribution will be not much different from any other "typical" number sequence of similar asymptotics). | |
| Jan 11, 2010 at 13:56 | comment | added | darij grinberg | Here is my attempt at explaining the many small primes: At first, it is clear that the denominator of $\displaystyle\sum\limits_{k=1}^{p-1}\frac{1}{k}\binom{2k}{k}$ has only primes <p in its factorization (actually, it divides (p-1)!). As for the numerator, we notice that if we denote $\displaystyle\sum\limits_{k=1}^{n-1}\frac{1}{k}\binom{2k}{k}$ by F(n), then F(p) is not only divisible by p, but also by all primes between p and 2p-1 (because F(p) is congruent to F(u) mod u for every prime u between p and 2p-1, since the sums F(p) and F(u) differ from each other only by some terms which ... | |
| Jan 11, 2010 at 3:09 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 2.5 |
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| Jan 11, 2010 at 3:05 | comment | added | Michael Lugo | Also the denominator tends to consist of primes less than p, and the numerator of primes between p and 2p, plus a few larger ones. There isn't necessarily one large factor; for example for p = 31 the numerator has prime factors 31^2*37^2*41*43*47*53^2*59*61*73*1801*6143. | |
| Jan 11, 2010 at 2:46 | history | answered | Mariano Suárez-Álvarez | CC BY-SA 2.5 |