       Re: Simple n-tuple problem - with no simple solution

• To: mathgroup at smc.vnet.net
• Subject: [mg115874] Re: Simple n-tuple problem - with no simple solution
• From: Dana DeLouis <dana.del at gmail.com>
• Date: Sun, 23 Jan 2011 17:32:03 -0500 (EST)

```On Jan 22, 3:23 am, DrMajorBob <btre... at austin.rr.com> wrote:
> Here are some timings.

> n = 5; addends = Rationalize@Range[0, 1.0, 0.05];
>
> {1.44372, 192}
> {23.6318, 192}

> n = 7;
> {1.85135, 364}

> n = 9;
> {2.04216, 488}

> The Compositions method rendered Mathematica unresponsive at n = 10, but
> here's n = 25 for my approach:
> n = 25;
>
> {2.57619, 627}
>
> I'd say Solve (which must use Integer Linear Programming for this) is the
> way to go.

Hi.  Just some thoughts.   With the smallest unit 1, the most n can be is 20.

If we assume that solution sizes smaller than n have implied 0's, then...

IntegerPartitions[20,5]//Length //Timing
{0.000085,192}

IntegerPartitions[20,20]//Length //Timing
{0.000244,627}

If we want to look at each exact n, then the function uses {n}

Table[Length[IntegerPartitions[20,{j}]],{j,20}]
{1,10,33,64,84,90,82,70,54,42,30,22,15,11,7,5,3,2,1,1}

Accumulate[%]
=
{1,11,44,108,192,282,364,434,488,530,560,582,597,608,615,620,623,625,626,627}

By adding them up, we see that for n=5, we get the 192.  For n=7, 364, etc.

If the question was "Count all possible combinations" as in the last total above, then one additional way...

Timing[SeriesCoefficient[
1/Times @@ (1 - n^#1 & ) /@
Range, {n, 0, 20}]]

{0.0003, 627}

= = = = = = = = = =
HTH  : >)
Dana DeLouis

```

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