# Can we represent this sequence as a function of n?

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. – joriki Oct 10 '12 at 14:45
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. – anon Oct 10 '12 at 14:46
oeis.org/… – Michael Albanese Oct 10 '12 at 14:47
oeis.org/A133411 – nav_jan Oct 10 '12 at 14:48
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. – Gerry Myerson Oct 17 '12 at 0:56

This is the unique minimum degree interpolated polynomial: $$-\frac{23}{15}n^5+\frac{175}{6}n^4-\frac{572}{3}n^3+\frac{3329}{6}n^2-\frac{3559}{5}n+324$$ whose values at $1,2,\ldots,6$ are $4,12,24,48,240,720$.

This can be found using GAP via:

InterpolatedPolynomial(Integers,[1,2,3,4,5,6],[4,12,24,48,240,720]);


While this is unlikely the answer you are looking for (since it has a negative leading coefficient and is of the maximum degree possible; moreover, your sequence appears to grow around $O(n!)$, whereas this polynomial grows as $O(n^5)$), you're probably not going to get a better answer without explaining what the sequence actually is.

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Don't you mean "is of the minimum degree possible"? – Daniel Hast Oct 17 '12 at 7:23
I mean, it's the maximum possible degree of a minimum degree interpolation polynomial for 6 points. If it turned out to have degree 2, for example, then there would be some redundancy in the data points (suggesting the polynomial is indeed correct). – Douglas S. Stones Oct 17 '12 at 7:54

I wrote out the factorizations of the first 24 numbers:

================================================================

#   A133411 (b-file synthesized from sequence entry)
2   3   5   7  11  13  17  19  23  29  31
-   -   -   -   -   -   -   -   -   -   -
1 1
2 2                    1
3 4                    2
4 12                   2   1
5 24                   3   1
6 48                   4   1
7 240                  4   1   1
8 720                  4   2   1
9 5040                 4   2   1   1
10 10080               5   2   1   1
11 20160               6   2   1   1
12 221760              6   2   1   1   1
13 665280              6   3   1   1   1
14 8648640             6   3   1   1   1   1
15 17297280            7   3   1   1   1   1
16 294053760           7   3   1   1   1   1   1
17 5587021440          7   3   1   1   1   1   1   1
18 27935107200         7   3   2   1   1   1   1   1
19 642507465600        7   3   2   1   1   1   1   1   1
20 1927522396800       7   4   2   1   1   1   1   1   1
21 13492656777600      7   4   2   2   1   1   1   1   1
22 26985313555200      8   4   2   2   1   1   1   1   1
23 782574093100800     8   4   2   2   1   1   1   1   1   1
24 24259796886124800   8   4   2   2   1   1   1   1   1   1   1
-   -   -   -   -   -   -   -   -   -   -
2   3   5   7  11  13  17  19  23  29  31


================================================================

There is no closed form expression for this. However, a more realistic goal is to have the ability to find the next larger element, $a_{n+1},$ in the list up to element $n,$ then the next one after that, and so on. This has been done. For general terminology, you want On Highly Composite Numbers by Jean-Louis Nicolas, pages 215-244 in a book called Ramanujan Revisited, edited by George Andrews et al, appearing in 1988. You may email me for a pdf. Then, the specific method is by his student Guy Robin, Methodes d'optimisation pour un probleme de theorie des nombres, R.A.I.R.O. Informatique theoretique 17, no. 3, 1983, pages 239-247.

The reason it is possible to do such a thing is that highly composite numbers bear some resemblance to superior highly composite numbers. S.H.C. numbers have the basic feature, nonincreasing exponents, but in addition the ratios of the exponents for distinct primes $p \neq q$ must approximate the outcome of Ramanujan's recipe. Between consecutive S.H.C. numbers, one may find all the h.c. numbers as a sort of operations research problem. This is all, well, difficult. So it all depends on why this is important to you, and how important. The short version, meanwhile, is that you calculate the s.h.c larger than you current $a_n,$ calculate the h.c. numbers up that high, and take the smallest one that is a multiple of $a_n.$ If you need to go up an extra s.h.c number, so be it.

Note that finding the first s.h.c number which is a multiple of your fixed number $a_n$ is entirely straightforward, as for any given $p^e,$ one may find the first (largest) $\delta > 0$ such that the s.h.c. number $N_\delta$ is divisible by $p^e.$ Do this for all primes that divide your fixed $a_n$ and there you have it.

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If we divide the terms by 4 we have $1,3,6,12,60,180...$ now we can proced to see a pattern, let's see the representation in prime numbers, $6=2\times 3$, $12=2^2\times 3$, $60=2^2\times 3 \times 5$ and $180=2^2\times 3^2 \times 5$, ...
So, taking $1=2^0$ and $3=2^0\times 3^1$ we have the list:

$2^0$
$2^0\times 3^1$
$2^1\times 3^1$
$2^2\times 3^1\times 5^0$
$2^2\times 3^1\times 5^1$
$2^2\times 3^2\times 5^1\times 7^0$

We see that the sum of the exponents of each line is $0,1,2,3,...$ When the sum of the exponents of each line is odd we add a new prime factor elevated to $0$. Then the next two number in the sequence is:
$2^2\times 3^2\times 5^1\times 7^1=1260$
$2^2\times 3^2\times 5^2\times 7^1\times 11^0=6300$

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Well, that's a pattern all right, but OP thinks it's A133411 at oeis.org --- does your pattern give that sequence? – Gerry Myerson Oct 17 '12 at 0:57
This pattern works with the number of the question but not with the complete list of oeis.org – José Ramírez Oct 17 '12 at 2:23
Where did the OP say it's A133411? The question indicates lack of mathematical maturity, which makes me think the question is unlikely to be about something as sophisticated as A133411. – Douglas S. Stones Oct 17 '12 at 21:32
@Douglas, OP didn't exactly say it's A133411, but OP did write "I have already seen that link but could not understand how to compute nth term from it" which I took, perhaps mistakenly, to mean that OP is interested in a formula for A133411. – Gerry Myerson Oct 18 '12 at 2:02