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Topic: Math Problem
Replies: 6   Last Post: Oct 13, 2008 5:38 PM

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Tsukino_Kaji

Posts: 3
Registered: 8/23/08
Math Problem
Posted: Aug 23, 2008 1:20 AM
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I wasn't sure were to put this, but I had it up in another forum and it is for recreation so I put it here.




Original post:
Ok so I haven't realy done any serious math in like 9 years and was wondering if there was any was to set this up as an equation;

So there'll be A, B and C. Every sequence A will produce one B and every B will produce one C. There will always be one base sequence and every A, B and C will give you another sequence.

1. Start with the bace sequence in which you obtain A(You never obtain another A and if you do god help you.).
2. You now have 2 sequences; Base+A.
First is A, B.
Second is A, B2, C.
3. You now have 5 sequnces; Base+A+ B2+C.
First is A, B3, C3.
Second is A, B4, C6.
Third is A, B5, C10.
Fourth is A, B6, C15.
Fifth is A, B7, C21.
4. You now have 30 sequences; Base+A+B7+C21.
Ending in A, B37, C671
5. You now have 710 sequences; Base+A+B37+C671

Problem is I run out of calculator space too fast and by hand is annoying and I am trying to find out how many I have on the 6th, 7th, 8th, 9th, and 10th run. Anything after that is moot realy, but humorous none-the-less just to show off the numbers. Because if you set off in 3, 4 will be deveastating and 5 it'll be over, but most likely it can finish until 6. Fun huh?

Only reply so far wrote:
This doesn't make sense to me. I'm not really sure what you're doing.

My reply was:
When you finish off the sequences, you start again with the new number of sequences. You always have 1, plus the number of As, Bs, And Cs you've acumulated. But when you add a B or C during a sequence set, it doesn't give you another sequence until the current set is over. Increasing the number of sequences in the set after it.
So first set is 2 sequences, The one you always have plus the one from A.
A produces a B in the first sequence.
Then In the second sequence the B that you now have produces a C and A produces another B.
You end the first set with an A, 2 Bs and a C making it so that you will have 5 sequences in the second set. The one you always have, plus A, Plus 2 Bs, plus a C.

I'm wondering if there's an easy way to calculate it out without have to just simply count every sequence in a set. As you can see set 6 now has 710 sequences and that's very tiring.



And that's how it stands at the moment, what do you all think?



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