User jose a rodriguez - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T07:02:05Z http://mathoverflow.net/feeds/user/21946 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90423/a-property-of-unimodal-sequences A property of unimodal sequences Jose A Rodriguez 2012-03-07T03:48:25Z 2012-05-06T01:14:34Z <p>It is well-known that $(-1)^j \sum_{i=0}^j (-1)^i\binom{n}{i} \geq 0$. This inequality can be used to prove Bonferroni's inequalities for example. Recently I noticed that a similar inequality applies to non-negative unimodal sequences with alternating zero sum: if $a_0 \leq a_1 \leq \ldots \leq a_k \geq a_{k+1} \geq \dots \geq a_{n}$ with $\sum_{i=0}^n (-1)^i a_i = 0$ then $(-1)^j \sum_{i=0}^j (-1)^i a_i \geq 0$ for $j=0 \ldots n$. Of course the binomial sequence is just a special case. My question is if anyone has any knowledge of this property of unimodal sequences having appeared or having been used anywhere in the literature? Thanks.</p> <p>Jose A Rodriguez</p> http://mathoverflow.net/questions/92753/sums-of-powers-mod-p Sums of powers mod p Jose A Rodriguez 2012-03-31T14:02:33Z 2012-04-03T04:51:20Z <p>For prime $p > 7$ with $p-1=rs$, $r>1$, $s>1$, let <code>$A=\{x^r|x \in \mathbb{Z}_p\}$</code> and <code>$B = \{x^s|x \in \mathbb{Z}_p\}$</code>. If $g$ is a primitive root mod $p$ then <code>$A = \{0\} \cup \{g^{ir}|0 \leq i &lt; s \}$</code> and <code>$B = \{0\} \cup \{g^{js}|0 \leq j &lt; r \}$</code>. Is it always true that $\mathbb{Z}_p \neq A + B$?</p> http://mathoverflow.net/questions/92753/sums-of-powers-mod-p/92959#92959 Comment by Jose A Rodriguez Jose A Rodriguez 2012-04-03T13:50:19Z 2012-04-03T13:50:19Z Very nice, it isn't obvious to see how to cast the problem into the form required by the theorem. http://mathoverflow.net/questions/92753/sums-of-powers-mod-p/92834#92834 Comment by Jose A Rodriguez Jose A Rodriguez 2012-04-02T17:17:51Z 2012-04-02T17:17:51Z Nice. Probably the most difficult case is when (r,s) = 1. http://mathoverflow.net/questions/92753/sums-of-powers-mod-p Comment by Jose A Rodriguez Jose A Rodriguez 2012-04-01T02:23:57Z 2012-04-01T02:23:57Z @Mark: Thanks, it would be interesting to check if my other observation ( that for fixed p and different factorizations of p-1 = rs with 1 &lt; r &lt;= s then as r increases, |A+B| decreases) holds as well. http://mathoverflow.net/questions/92753/sums-of-powers-mod-p Comment by Jose A Rodriguez Jose A Rodriguez 2012-03-31T18:33:38Z 2012-03-31T18:33:38Z Yes, Seva, I agree. How do we quantify the number of duplicate sums? http://mathoverflow.net/questions/92753/sums-of-powers-mod-p Comment by Jose A Rodriguez Jose A Rodriguez 2012-03-31T18:12:42Z 2012-03-31T18:12:42Z @Felipe: Yes, so far it appears to hold numerically. I just tried p = 113117 for fun. p-1 = 4*28279. When r = 2, s= 2*28279 I get |A+B| = 98903. If r = 4, s = 28279 then |A+B| = 86330. The gap between A+B and Z_p seems to grow larger and larger as p increases. Moreover it also appears that for given p and different factorizations of p-1 = rs, as r and s get closer, |A+B| gets smaller, so the largest |A+B| seems to occur when r = 2, s = (p-1)/2. Maple is just too slow, I need to write a program to test larger cases and see a trend. @Zack: Forgive my ignorance, how do you get that?