I was wondering the function for getting the number of combinations of 0 1 2, repeatable, but with any combination containing a 02 disallowed.(adsbygoogle = window.adsbygoogle || []).push({});

Up to four places the sequence of answers is 3, 8, 21, 55 that I have counted and verified, It would seem the next few are 144, 377, 987, 2584, I have not verified those for obvious reasons.

The system I worked out to get those numbers is that if, once up to two digits, you name duples as a-h skiping '02' i.e. a='00', b='01',(02 is not allowed) c='10'...,h = '22' and then consider them to be overlapping, so that the second digit of one determines the first digit of the other.

The 8 for two places is obvious as there are 8 duples, that 8 break down to 3(a,c,f) that can only have 2(a,b) that follow, 3(b,d,g) that can have 3 follow(c,d,e) and 2(e,h) that can have 3 follow(f,g,h).

So (3*2)+(3*3)+(2*3) = 6+9+6 = 21

The 21 then breaks down to 8,8,5

(8*2)+(8*3)+(5*3) = 55,

55 breaks down into 21,21,13

(21*2)+(21*3)+(13*3) = 144

144 breaks down into 55,55,34 (The pattern here repeats in that the first two are the answer from before(edit: the answer from two before), and the last one is the two from before added together(13 = 8+5),(34 = 21+13)

But alas, I don't know how to put such a recursive formula into terms that I can supply the number of digits to get the answer without starting at a known point and continuing to the number of digits I want. However, I believe such a formula should exist? One should employ some sort of (n+1) system or something I imagine, but my skills there some short I don't even know where to start.

Finally, to put some background to that, I am figuring the number of possible rhythms in 8 measures of 2/4 going only to 16th notes. So far it would seem the answer is close to maybe 4*10^27. I figure that if 1 billion people wrote 150/minute, it would take almost 14 billion years to accumulate them all.

Also, if you have 9 gallons of water, there are as many rhythms as there are molecules of water. neat stuff

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# Combinatorial Question

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