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In the following sequence of fractions,

$$21/8,273/32,1333/72,4161/128,10101/200,20881/288$$ the denominators are simply $8n^2$, but can't figure out what the numerator is. It's not in OEIS.

Also need help with this one if possible: $$1/2,73/18,601/50,2353/98,6481/162$$


EDIT: Robert and Brian yes you are right. In the second sequence I had the third and fourth terms wrong. They are now corrected. I double checked the other ones too. Apologies.

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What leads you to think there should be a pattern in the numerators? I.e. what is the context of this question? – peoplepower Sep 3 '12 at 6:01
Looks like values of a quartic polynomial by a higher order difference test - the fourth iterated differences are constant 384. Granted,only two samples remain at point:-) – Jyrki Lahtonen Sep 3 '12 at 6:02
The second one seems not to be so simple. Are you sure the denominator in the third term is $40$ rather than $50$? – Robert Israel Sep 3 '12 at 6:29
@RobertIsrael:Yep, plus some other mistakes. See edit in OP above. – ben Sep 3 '12 at 6:47
up vote 6 down vote accepted


Here's how. Let $f(n)$ be the sequence of numerators. Define the higher order differences recursively $\Delta^{[1]}f(n):=\Delta f(n):=f(n+1)-f(n)$, and for all $k>1$ let $\Delta^{[k+1]}f(n):=\Delta(\Delta^{[k]}f)(n).$ This is the discrete version of higher derivatives.

For the sequence of numerators $f(n)$ we get that $\Delta^{[4]}f(n)$ is the constant $384$ (but we can only compute it for $n=1,2$). Because $D^4x^4=24$, and $384=16\cdot24$ we recognize this as $\Delta^{[4]}(16n^4)$. For the sequence $g(n):=f(n)-16n^4$ we repeat the exercise, and this time we notice that already the second order differences $\Delta^{[2]}g(n)$ form a constant sequence $8$. This is the second derivative of $4x^2$, and gives the second term. The remaining term of the sequence, $g(n)-4n^2$, is just constant $1$.

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This is fantastic. Following your formula I've been able to reproduce the $384$ that results from the fourth finite differencing, but what is $D^4x^4$? – ben Sep 3 '12 at 17:07
Good! Sorry, I skimmed that part. By $D^4x^4$ I simply meant the fourth derivative of the polynomial $x^4$. It is an easy exercise that for any polynomial $p(x)$, the leading terms of $D^kp(x)$ and $\Delta^{[k]}p(n)$ agree. That's why I always calculated the higher order differences until they appeared to formed a constant function, so that the leading term is all there is. Effectively this is a method of building a polynomial representation of any sequence term-by-term, starting from the highest degree term. – Jyrki Lahtonen Sep 3 '12 at 17:40
Got it. Interesting that the fourth differencing of the second set of numerators is also $384$, which points to an interesting relation between the two sequences. I would not have noticed this had you not explained the differencing method. Moreover, your approach really takes the guesswork out of this kind of problem. So much so that you can write a matlab algorithm based on it. In this way with one or two clicks I found that the second sequence of numerators is $$16n^4-32n^3+20n^2-4n+1$$ which is the same result as Brian's answer in expanded form. – ben Sep 4 '12 at 2:38

The numerators are $21=5^2-2^2$, $273=17^2-4^2$, $1333=37^2-6^2$, $4161=65^2-8^2$, $10101=101^2-10^2$, and $20881=145^2-12^2$. A little work with finite differences shows that the sequence $5,17,37,65,101,145$ is given by $4n^2+1$, so it appears that your $n$-th term is $$\frac{(4n^2+1)^2-(2n)^2}{8n^2}=\frac{(4n^2-2n+1)(4n^2+2n+1)}{8n^2}\;,$$ or, if you prefer, $$\frac{16n^4+4n^2+1}{8n^2}\;.$$

Added: In the second sequence there may be a typo in the third term: it may have been supposed to be $2353/98$. With that emendation the numerator of the $n$-th term would be $2n(n-1)d_n+1$, where $d_n$ is the denominator of the $n$-th term. Of course that still leaves the denominators a bit of a puzzle: $d_n=2(2n-1)^2$ would work except for the $40$, which ought by that rule to be $50$. Before spending more time on it, I’d want to be sure that the sequence has been given correctly.

Added2: The corrected sequence, with $601/50$ as third term, fits the pattern that I suspected: the $n$-th term is $$\frac{4n(n-1)(2n-1)^2+1}{2(2n-1)^2}=2n(n-1)+\frac1{2(2n-1)^2}\;.$$

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Yes, sorry about that. See the edit to the OP above. – ben Sep 3 '12 at 6:48
This answer is also excellent, insightful, and addresses both sequences. But as they say "give a man a fish and he eats for a day, teach a man to fish and he eats every day." Jyrki's answer has taught me how to fish. Nonetheless I will give this answer a well-deserved upvote. Thanks! – ben Sep 4 '12 at 2:44

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