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I have this sequence : 4,12,24,48,240,720,....

I order to find nth term of this sequence , can we represent this sequence as a function of n.

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    A sequence is by definition a function whose argument it a natural number. Perhaps you mean something like a formula in closed form? That has nothing to do with functions.2012-10-10
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    Given a finite sequence, there are infinitely many different ways of extending the sequence (alternately: infinitely many sequences that start with the given initial subsequence), so it will often be ambiguous what "sequence" we are actually referring to. Even if we restrict to sequences with "closed-forms." The successive divisors here seem to be 3,2,2,5,3, but without more terms I don't think I can't make out a pattern.2012-10-10
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    http://oeis.org/search?q=4%2C12%2C24%2C48%2C240%2C720&language=english&go=Search2012-10-10
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    http://oeis.org/A1334112012-10-10
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    @MichaelAlbanese I have already seen that link but could not understand how to compute nth term from it.2012-10-10
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    @joriki yes you are right I want something in form of n so that I can easily compute the nth term Otis series. Thanks2012-10-10
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    I can't believe, that even after 2 days you didn't find anything more trying it yourself.2012-10-12
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    So where does this come from @nav_jan? Do you know for sure that the OEIS sequence is the one you want the closed form for? There isn't enough information here for us to help you.2012-10-13
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    @GottfriedHelms I am here for help, your comment did not help at all. Thanks.2012-10-13
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    If this is A133411, then I doubt a closed-form formula as a function of $n$ exists. If you look at the sequence of quotients, it's a jumbled list of primes, without much pattern as to when the next new prime comes in or when you repeat an old one, or which old one you repeat.2012-10-17
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    This is hard because it is the intersection between two sets-the highly composite numbers and the multiples of previous elements. I suspect the ratio between terms will be the greedy algorithm. For a prime $p$ where the highest power dividing the current number is $p^n$, the increase in the ln of the number of factors will be $\ln (\frac {n+2}{n+1})=\ln (1+\frac 1{n+2})\approx \frac 1{n+2}$, so the prime chosen will the prime that minimizes $\frac {\ln p}{n+2}$. I haven't checked that this reproduces the series so far.2012-10-17
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    @RossMillikan, that particular description of greedy does not work. Assuming you are taking a prime not currently present as $p^0,$ the resulting sequence is just the powers of $2.$ The idea that there is some greedy algorithm that works is certainly interesting.2012-10-17
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    @WillJagy: I think it does. Adding in a prime not already there doubles the number of factors, so increases the log of the number of factors by $\ln 2$2012-10-17
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    @RossMillikan, it appears I have misunderstood your idea. Meanwhile, I took the list of the first 24 numbers in this sequence and pasted those into my answer, with the prime factorization of each indicated simply by the exponents for the primes 2 through 31.2012-10-17

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