Sequence
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the following sequence
1,2,4,5,7,9,10,12,14,16,17
has one odd number followed by 2 evens, then 3 odds, 4 evens, and so on. what number is th 2003th term?
i hate sequence questions
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hmm..
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Okay, google did the trick, dun think I'll be able to think this on my own.
http://mathworld.wolfram.com/ConnellSequence.html -
2n-1 doesnt seem to fulfil the third and second term
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if its as simple as that, i would have figured it out a long timeOriginally posted by NewAge:2n-1 doesnt seem to fulfil the third and second term
hell, i tried figuring out the formula for the sequence of even and odd, formula for the 'x-th' term for odd numbers, formula for the 'x-th' term for even number, and some other weird stuff -
this link is much betterOriginally posted by CoolMyth: -
can explain this part?
"Suppose we reach the number k^2 in the sequence. This will happen at
the kth triangular number, i.e. at at index of k(k+1)/2. For example,
25 = 5 x 5 has k = 5, so it is the 5(5+1)/2 = 15th number in the
sequence. " -
You can see that the first column forms the k^2 numbers. So I guess it means the 5th row bah.Originally posted by mystiv:can explain this part?
"Suppose we reach the number k^2 in the sequence. This will happen at
the kth triangular number, i.e. at at index of k(k+1)/2. For example,
25 = 5 x 5 has k = 5, so it is the 5(5+1)/2 = 15th number in the
sequence. "