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I have the following sequence:

0 0 2 6 12 20 30 42 56

which can be summed as: $$ 2 \sum_{i=0} {(n_i/2 - 0.5) n_i}. $$ I also have the first 9 terms (i.e., $n=0,1,2,\ldots,8$) of two (unrelated) sequences, neither of which appears geometric, and both of which should begin with 2*:

 0 2 4 10 20 34 52 74
 0 2 4 6 16 34 60 94

I would be grateful if anyone could help me find a series for each of these, or at least give some advice as to how I should tackle it.

Thanks.

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You might look at the Online Encyclopedia of Integer Sequences (OEIS). The first sequence looks like $k(k+1)$, easily summed if you wish. There is some noise at the beginning of the other two, but soon first differences look as if they grow linearly, so the sequences may be more or less quadratics. One could only know for sure if one knew context. –  André Nicolas Apr 28 '12 at 15:48
    
What are the $n_i$ supposed to be? –  J. M. Apr 28 '12 at 16:03
    
I have figured out the first of the two sequences, which can be written as $\sum_{i=1}(2(n_i - 2 + 2/n_i)*n_i$, but I'm still struggling with the second. The $n_i$ are dividing cells, which may undergo several rounds of asymmetric self-renewal before differentiating, which is why the first two numbers in the sequence are zero (no differentiated cells generated in cycles 0 or 1). I think the first zero can be lopped off, so the second sequence will read: 0 2 4 6 16 34 60 94. –  eric Apr 28 '12 at 16:26
    
What exactly do you want to know? The pattern of those two sequences? There are dozens of sequences starting with those $8$ numbers. –  TMM Apr 28 '12 at 17:53
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"Even if there are dozens, I'd be grateful to learn some or all of them, whose predictions I can test against my data." - Given any $8$ numbers, one can find infinitely many formulas describing a different sequence but matching those numbers. In fact, given $1, 1, 1, 1, 1, 1, 1, 1$, I can easily give you a formula describing a sequence that continues with $\pi, e, 100, \sqrt{3}$. Would that sequence be the sequence you are looking for? Probably not. –  TMM Apr 28 '12 at 23:37

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