<p>Can anyone please solve this sequence and give me the formula?
0,1,1,2,2,3,3,4... every positive integer is repeated.</p>
<p>If it's any help, I got as far as n/2 for even numbers and (n/2)-(1/2) for odd numbers? And every successive number differs by +/- 1/2, as in (n/2)-(1/2), (n/2)-(-1/2)+(1/2), (n/2)-(1/2)+(1/2)-(1/2) and so on.</p>
<p>the sum of the integers is n(n+1)/2. Since there are two of each, it is n(n+1). Is this finite, if so you would just plug in the last integer value for n. If it is infinite, I can’t think of a way to solve this for sure, but it seems that it diverges, making the answer infinity.</p>
<p>0 1 1 2 2<br>
1st term 2nd term 3rd term 4th term 5th term
(odd term) (even term) (odd term) (even term) (odd term)</p>
<p>Notice that for a specific number, the following sequence on the term’s number happens: “even, odd”.</p>
<p>The first time the number 1 appears is in the 2nd position. If you divide 2 by 2, you have 1, which is the respective number. As we figured out the sequence of “even, odd”, the 3rd term is also 1.</p>
<p>Suppose you want to find out what the 13th term is.</p>
<p>13 is odd, so since a number appears for the first time as an even term, the 12th term should be the same as the 13th. Using the same methodology from the “1 example” above, take 12/2 and you have 6. This means that the 12th term is 6 and the following term, the 13th, is also 6.</p>
<p>As you said, there are different formulas to find the terms depending on whether the specific term is odd or oven. Thus, as far as I know, you can’t find an equation for the sequence as a whole. This could be the sum of two arithmetic progressions, if you want.</p>
<p>Maybe someone will come with a better answer, let’s wait.</p>
<p>a(sub n+1) = -a(sub n) + n +1 [For n > 1]</p>
<p>^isn’t that recursively defined? I thought the OP was looking for a non-recursive one. The best I can find is 2 formulas for the series, can’t seem to find a general one.</p>
<p>A recursive formula would do, but for this equation: a(sub n+1)= -a(sub n) + n +1 [For n > 1], you would have to assume zero is given as a(sub n), which it really isn’t. Plus, even when you assume that a(sub n) IS zero, and that substitute values, you get: -0+2+1, and you get 3. So this doesn’t apply either.
Wong tong tong, even I can only find two formulas for the series, not a general one.</p>
<p>And Jalmoreno, it doesn’t converge. It diverges.</p>
<p>Whoops, sorry. Didn’t catch that. And yeah, what I posted doesn’t work for 0 (the term where n=1).</p>
<p>Formula is ([-1+{-1}^n]-2)(n/2)+(n/2+1)([1+{-1}^n]/2)</p>
<p>^^^ that’s how you have to define it since you need separate formulas for even and odd terms.</p>
<p>Does that seem right?</p>
<p>I got it. Its actually floor(n/2) :)</p>
<p>Another question (sorry, I have a calculus exam tomorrow and there’s a lot of last minute cramming):
I need to make this converge to 1/2,</p>
<p>((n^2 - n)^1/2) - n</p>
<p>What I’m doing is completing the square in the underroot. So I get ((n-1/2)^2 - (1/2)^2)^1/2 -n. I then discard the (1/2)^2 because I’m taking (n-1/2)^2 to be equal to infinity and (1/2)^2 becomes negligible in comparison. Is this legit? As in, does this stand? Because I can’t find any other way to make it converge to 1/2.</p>
<p>cos^2(pi<em>n/2)</em>n/2+sin^2(pi<em>n/2)</em>(n-1)/2</p>
<p>hope this will be helpful</p>
<p>[2n - 1 - (-1)^(n+1)] / 4</p>
<p>My logic:</p>
<p>n minus a-sub-n is always [n/2 + (either 1/2 for odd n’s, or 0 for even n’s)].</p>
<p>The formula for alternating 1/2’s and 0’s is (1 - (-1)^(n+1)) / 4.</p>
<p>So the the formula is n - (n/2) - [(1 - (-1)^(n+1)) / 4], which simplifies to what I typed above.</p>
<p>S<em>1 = 0
S</em>2 = 1
S<em>3 = 2
S</em>4 = 4
S_5 = 6, etc.</p>
<p>It’s pretty easy to prove by induction that S<em>(2k) = k^2. Also, S</em>(2k+1) = k^2 + k.</p>
<p>I know the thread is over a year old, but whatever :)</p>