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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.

  • 0
    @RossMillik$a$n, 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

3 Answers 3

4

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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    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).2012-10-17
4

I wrote out the factorizations of the first 24 numbers:

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#   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 

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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.

3

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:
$2^2\times 3^2\times 5^1\times 7^1=1260$
$2^2\times 3^2\times 5^2\times 7^1\times 11^0=6300$

  • 0
    @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 $f$rom it" which I took, perhaps mistakenly, to mean that OP is interested in a formula for A133411.2012-10-18