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Ghost Post
Icarian Member

 Posted: Sun Mar 31, 2002 8:47 pm    Post subject: 1 ... 10, 12, 15, 19, 24, 31, 41, 55, 74, 100 ... For fun, can you work out the beginning of the sequence too?
i_h8_evil_stuff
Daedalian Member

 Posted: Sun Mar 31, 2002 9:01 pm    Post subject: 2 well, it could be (the number 9 spaces before it)^2, which would be: ...10^(1/2), 2(3^(1/2)), 15^(1/2), 19^(1/2), 2(6^(1/2)), 31^(1/2), 41^(1/2), 55^(1/2), 74^(1/2), 10, 12, 15, 19, 24, 31, 41, 55, 74, 100, 144, 225, 361, 576, 961, 1681, 3025, 5476, 10000... ...but i dont think thats what u meant ------------------ The name explains it all.
mith
Pitbull of Truth

 Posted: Sun Mar 31, 2002 11:40 pm    Post subject: 3 ... -6, 16, -5, 11, 0, 11, 1, 7, 2, 8, 4, 6, 3, 6, 5, 6, 4, 5, 5, 6, 5, 5, 5, 6, 6, 6, 6, 7, 8, 9, 10, 12, 15, 19, 24, 31, 41, 55, 74, 100, 136, 186, 255, 350, 481, 662, 912, 1257, 1733, 2390, 3297... If that isn't right, I don't have a clue
Ghost Post
Icarian Member

 Posted: Mon Apr 01, 2002 12:17 am    Post subject: 4 The funny thing is, even though I have not a clue how your sequence works Mith, you have one of the two correct numbers that follow 100. I'd be interested to see your reasoning. Oh ... did I forget to say that there were two different numbers that follow 100 depending on how you solve this
cubestudent
3D Member

 Posted: Mon Apr 01, 2002 1:10 am    Post subject: 5 y'know, i got the same answer as mith, except i have no idea how he went further back than the 6. *sigh* too bad it's wrong. er. i do know. gah, i'm an idiot. [This message has been edited by cubestudent (edited 03-31-2002 08:46 PM).]
cubestudent
3D Member

 Posted: Mon Apr 01, 2002 1:45 am    Post subject: 6 oh, yeah, and: code:``` the pattern we have is: S(1) = 10; S(2) = 12; S(3) = 15; S(4) = 19; S(5) = 24; S(6) = 31; S(7) = 41; and: for all x | x > 7, S(x) = S(x-1) + S(x-3) - S(x-7) this gives us S(8) = S(7)+S(5)-S(1) = 41+24-10 = 55 and S(9) = S(8)+S(6)-S(2) = 55+31-12 = 74 and S(10) = S(9)+S(7)-S(3) = 74+41-15 = 100 which confirm your sequence and S(11) = S(10)+S(8)-S(4) = 100+55-19 = 136 and S(12) = S(11)+S(9)-S(5) = 136+74-24 = 186 and S(13) = S(12)+S(10)=S(6) = 186+100-31 = 255 which we (mith much quicker than i) guessed were the next couple ```
Ghost Post
Icarian Member

 Posted: Mon Apr 01, 2002 2:05 am    Post subject: 7 Whatever works works. We all know that the answer to these things depends on how you do them. As long as your answer is correct (which it appears to be) then its correct. In my sequence which goes 100, 136 ... the next number is 185. btw... 127 is the other number I thought someone might come up with. p.s. this might help you get mine ... 100 is the only point at which you may get two alternative answers using my methodology. [This message has been edited by quidnunc (edited 03-31-2002 09:06 PM).]
mith
Pitbull of Truth

 Posted: Mon Apr 01, 2002 3:14 am    Post subject: 8 actually, though i'm sure it ends up the same, the way i did it was this: taking each pair of consecutive numbers, the differences are: 2, 3, 4, 5, 7, 10, 14, 19, 26 Notice that: 7 = 5+2 10 = 7+3 14 = 10+4 etc. So, using *that* recursive function, which works backwards as well, I came up with the sequence above. (I prefer this because the function only uses two previous terms, and the seed numbers aren't so random (10, and 2, 3, 4, 5))
cubestudent
3D Member

 Posted: Mon Apr 01, 2002 4:13 am    Post subject: 9 fine, mith. if you're looking for elegance, try this on for size, punk : this is a pyramid of the differences, and differences of differences, etc. with the bottom row being the sequence in question, and the bolded being what we were given. code:``` 1 1 1 1 2... 1 1 1 1 2 3 4 5... 1 1 1 2 3 4 5 7 10 14 19... 2 3 4 5 7 10 14 19 26 36 50... 10, 12, 15, 19, 24, 31, 41, 55, 74, 100, 136, 186... ``` you'll notice that the layers of differences are the same sequence, just disjoint. and the resulting sequence we were given is the same sequence, too, except with 5 added to each term. :P [This message has been edited by cubestudent (edited 03-31-2002 11:32 PM).]
Ghost Post
Icarian Member

 Posted: Mon Apr 01, 2002 4:30 am    Post subject: 10 cubestudent, using your pyramid its a wonder you can't get my sequence - your eventual pattern of numbers is 1 1 1 1 2 3 4 5 7 10 ... Now where have I seen them before? [This message has been edited by quidnunc (edited 03-31-2002 11:31 PM).]
cubestudent
3D Member

 Posted: Mon Apr 01, 2002 5:25 am    Post subject: 11 i still don't get it. i really wish i did; it's one of a very interesting set of sequences: the first of which is the powers of two: S(x) = S(x-1) + S(x-1) ; ; S(1) = 1 yielding: 1, 2, 4, 8, 16, 32, 64, 128, . . . and the second of which is Avogadro's: S(x) = S(x-1) + S(x-2) ; ; S(1) = S(2) = 1 yielding: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . . the third i don't recognize: S(x) = S(x-1) + S(x-3) ; ; S(1) = S(2) = S(3) = 1 yielding: 1, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, . . . and the fourth is yours (after subtracting 5 from ea. term): S(x) = S(x-1) + S(x-4) ; ; S(1) = S(2) = S(3) = S(4) = 1 yielding: 1, 1, 1, 1, 2, 3, 4, 5, 7, 10, 14, 19, 28, . . . tell me, tell me, tell me what i should recognize about it!
Ghost Post
Icarian Member

 Posted: Mon Apr 01, 2002 5:28 am    Post subject: 12 Look at the integer value of n/10 in the numbers I gave.
cubestudent
3D Member

 Posted: Mon Apr 01, 2002 6:14 am    Post subject: 13 oooh! very cool! and by cool i mean totally sweet!
mith
Pitbull of Truth

 Posted: Mon Apr 01, 2002 3:12 pm    Post subject: 14 um... but, that stops working after 100, doesn't it? i hope that doesn't have anything to do with the answer.
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