Re: Inequalities or histogram or something.......

*To*: mathgroup at smc.vnet.net*Subject*: [mg89346] Re: [mg89338] Inequalities or histogram or something.......*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Sat, 7 Jun 2008 02:51:25 -0400 (EDT)*References*: <200806061047.GAA24209@smc.vnet.net>

I don't have a general formula, and having given it only a little thought, am not convinced an explicit formula can be found in the general case. But I may be wrong. But here are some observations. First, note that in your post you make the assumption that 1<=k but then you give an example with k==0. Presumably you mean k>=0? Also, your assumption k<v follows from 1<= x1 < x2 < x3 <= v and k = x1 + x3 - x2 - 2, since clearly k==x3 -(x2-x1)-2 must be less than x3 which is less than v. But this is not important. It also easy to see that your equation can only have solutions if v>k +2, which of course is needed for your relation d(v,k) + d(v,v-k-3) = 2(k+1)(v-k-2) to make sense. The case v=3 then k must be 0 so we can assume that v>=4. Next, it is easy to prove that d[v,0]== v -2 The proof goes as follows. For k= 0 your equation becomes x1 + x3 - x2 - 2==0 which can be re-written as x1 = 2 -(x3-x2) Now, since x3-x2>0 and x1>=1, this only leaves the possibility that x1 = 1. So the equation becomes x3-x2=1. So d[v,0] is the number of ways of choosing two successive integers between 1 and v. You can choose any integer z such that 2<=z<=v-1 as the smaller one, and v+1 as the larger one. There are v-2 such choices. So now that we know that d[v,0]=v-2 and can write a reasonably fast program to find other values of d[v,k]. d[v_, 0] := v - 2 d[v_, k_] /; (v - 3)/2 < k < v - 2 := d[v, k] = 2 (k + 1) (v - k - 2) - d[v, v - k - 3] d[v_, k_] /; 0 < k <= (v - 3)/2 := d[v, k] = Length[Union[ Reduce[1 <= x1 < x2 < x3 <= v && k == x1 + x3 - x2 - 2, {x1, x2, x3}, Integers]]] This seems quite fast; for example d[10000, 2305] // Timing {0.283073, 4} d[10000, 7692] // Timing {0.000065, 35480112} So one could try to first explore this empirically, guess the formula and then it should not be hard to prove it inductively. But this is as much time as I can devote to this problem. Andrzej Kozlowski On 6 Jun 2008, at 19:47, Steve Gray wrote: > Can Mathematica help with this? Or can someone? > > I have positive integers x1,x2,x3,k,v. > > There are assumptions: > 1 <= k < v and > 1 <= x1 < x2 < x3 <= v. > > There is one equation: > k = x1 + x3 - x2 - 2. > > I need a symbolic solution for the number of combinations of x1,x2,x3 > that satisfy the equation under the assumptions. > > This will be a histogram of k vs. the number of solutions. > One numeric point on the histo: if v=8 and k=0, there are 6 solutions, > x1,x2,x3 = 1,2,3; 1,3,4; 1,4,5; 1,5,6; 1,6,7; 1,7,8; none with x1 > 1. > > I need a general symbolic solution in terms of D(v,k). I need a way to > show its derivation for a paper. I happen to know that > D(v,k) + D(v,v-k-3) = 2(k+1)(v-k-2). > > This is not overwhelmingly complicated but I don't know a decent way > to go about it. Thank you for any help, using Mathematica or not. (I > have > version 6.) > > Steve Gray >

**References**:**Inequalities or histogram or something.......***From:*Steve Gray <stevebg@roadrunner.com>