Date: 06/12/2003 at 23:40:49 From: Rachelle Subject: Number sequence The following sequence 1,2,4,5,7,9,10,12,14,16,17... has one odd number followed by two evens, then three odds, four evens, and so on. What number is the 2003th term?
Date: 06/16/2003 at 14:19:58 From: Doctor Douglas Subject: Re: Number sequence Hi Rachelle, For this problem, it's a lot easier to see what happens if we format the sequence in a triangle: 1 2 5 10 17 . . . 4 7 12 19 9 14 21 16 23 25 . . . You can see that each diagonal is either all even or all odd, and that they alternate. Also, notice that the numbers in the first column are all perfect squares (this observation will save us a lot of algebraic work). 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. We need the smallest triangular number bigger than 2003. We can estimate the k for which this happens by considering sqrt(2003*2) = sqrt(4006) = 63.29. 63 x 64/2 = 2016 62 x 63/2 = 1953 This tells us that the 2003rd number in the sequence is an ODD number somewhere between 62^2 = 3844 and 63^2 = 3969. We know that 3969 is the 2016th number in the sequence. Since we want the 2003rd, we know that it is 13 members before 3969, or 3969 - 13 x 2 = 3969 - 26 = 3943. Incidentally, this sequence is sometimes known as the "Connell sequence." There is even an explicit formula for the nth term: a(n) = 2n - Floor[(1 + Sqrt(8n - 7))/2] The notation "Floor(x)" means take x and round down to the next integer. If you plug n = 2003 into this, you'll obtain the answer that we found above. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/
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