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Certain and Curious Number Sequence

by Feryll
Tags: curious, number, sequence
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Feryll
#1
Dec2-11, 06:19 PM
P: 7
This is the number sequence a(x) whose output is determined by the greatest integer divisor n of any factorization of a with the additional rule that if an element is repeated in the factorization, the factorization must be thrown out.

EX: a(56). The factorizations are 1*56, 2*28, 4*14, 7*8, 7*2*4. Any of these are valid factorizations, take the greatest element of each and you have the output elements 56, 28, 14, 8, 7.

This sequence is consequential to the series below:
[itex]∏_{n=1}^T(1+n^{-1})[/itex]

Which is:

(1+1)(1+1/2)(1+1/3)(1+1/4)...

Now obviously, this function boils down to T+1, but we don't want to leave it at that. If you manually multiply out the terms, you'll get, for example for t=4, this:

1+1+1/2+1/2+1/3+1/3+1/4+1/4+1/6+1/6+1/8+1/8+1/12+1/12+1/24+1/24=5


As you see, the expansion of the series for term T is equivalent to adding all of the reciprocals of the elements of the function on a graph of our curious number sequence graph below or on the line y=T, and then multiplying the result by 2, due to how we defined the function in the first place.

What I'm asking for is "where is there more information and studies on this (relatively minimalistic) function?", although any further communal observations would be equally helpful. Here is the real meat of what I've been able to casually glean with my modest abilities:

1. No elements exists above the line of y=x.
2. No elements exist below the line of y=infa(x) where infa(x) is the inverse of the factorial (sorry, I don't know my terminology).
3. The addition of the reciprocals of the elements' x-values on any y-level with y being any natural number above 1 is 1/2.
4. The elements' x-values, with sole elements (thus existing on the line y=x), are all either primes or square primes, and it includes each and every prime and square prime.
5. I have a couple of results for calculating the elements existing on a line y=x/k for any natural number k with relation to the values existing on y=x (ie the primes and the squares of primes), although it is only a primitive algorithm which produces a subset of the elements.

I would create and display a graph, but I can't really capture what I'm trying to show here on such a surface. I will try to if someone asks for one.

I apologize in advance for any terminology or drunken logic errors, as I don't communicate with other math people very often, and my study in these areas are solitary.
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ramsey2879
#2
Dec3-11, 06:33 PM
P: 894
Quote Quote by Feryll View Post
This is the number sequence a(x) whose output is determined by the greatest integer divisor n of any factorization of a with the additional rule that if an element is repeated in the factorization, the factorization must be thrown out.

EX: a(56). The factorizations are 1*56, 2*28, 4*14, 7*8, 7*2*4. Any of these are valid factorizations, take the greatest element of each and you have the output elements 56, 28, 14, 8, 7.

This sequence is consequential to the series below:
[itex]∏_{n=1}^T(1+n^{-1})[/itex]

Which is:

(1+1)(1+1/2)(1+1/3)(1+1/4)...

Now obviously, this function boils down to T+1, but we don't want to leave it at that. If you manually multiply out the terms, you'll get, for example for t=4, this:

1+1+1/2+1/2+1/3+1/3+1/4+1/4+1/6+1/6+1/8+1/8+1/12+1/12+1/24+1/24=5


As you see, the expansion of the series for term T is equivalent to adding all of the reciprocals of the elements of the function on a graph of our curious number sequence graph below or on the line y=T, and then multiplying the result by 2, due to how we defined the function in the first place.

What I'm asking for is "where is there more information and studies on this (relatively minimalistic) function?", although any further communal observations would be equally helpful. Here is the real meat of what I've been able to casually glean with my modest abilities:

1. No elements exists above the line of y=x.
2. No elements exist below the line of y=infa(x) where infa(x) is the inverse of the factorial (sorry, I don't know my terminology).
3. The addition of the reciprocals of the elements' x-values on any y-level with y being any natural number above 1 is 1/2.
4. The elements' x-values, with sole elements (thus existing on the line y=x), are all either primes or square primes, and it includes each and every prime and square prime.
5. I have a couple of results for calculating the elements existing on a line y=x/k for any natural number k with relation to the values existing on y=x (ie the primes and the squares of primes), although it is only a primitive algorithm which produces a subset of the elements.

I would create and display a graph, but I can't really capture what I'm trying to show here on such a surface. I will try to if someone asks for one.

I apologize in advance for any terminology or drunken logic errors, as I don't communicate with other math people very often, and my study in these areas are solitary.
You present 2 interesting sequences, but how is one consequential to the other? I can't seem to be able to read between the lines!
Feryll
#3
Dec3-11, 08:36 PM
P: 7
Quote Quote by ramsey2879 View Post
You present 2 interesting sequences, but how is one consequential to the other? I can't seem to be able to read between the lines!
EDIT: I just noticed something on reflection that, being me, I got wrong in the OP. If you end up with two or more repeats of output elements when figuring out the factorizations' greatest factors, you must repeat it that many times, which I didn't really specify. See below for implciations.

Thanks for the response, ramsey!

If you take that product series and examine how, up to any term, how you're going to manually multiply it out and what you're going to end up with, you should be able to see it. Look at this similar series:

[itex]∏_{n=2}^T (1+n^{-1})[/itex]
(1+1/2)(1+1/3)(1+1/4)...

Under these circumstances (the original power series /2), you add one (multiply out all ones, only), you add up the reciprocals all natural numbers up to and including T (pluck one n^-1 out, then all 1s), then all the reciprocals of the product of any two natural numbers up to and including T (pluck 2 n^-1s out, then all 1s), then all the reciprocals of the product of any three natural numbers up to and including T (pluck 3 n^-1s out, then all 1s), and so on, which is standard FOILing/associative law application.

And here you can connect it to the sequence.

Note that you will never be able to multiply out multiple identical n^-1s, so that is the reason that, in the original sequence, that if you end up with a factorization that contains two of the same factors, you must throw it out.

Now, for example, a(84) has the factorizations 2*2*3*7, 4*3*7, 2*6*7, 2*3*14, 2*2*21, 4*21, 7*12, 3*28, 2*42, 1*84. I've eliminated the illegal factorizations, as they have repeating factors, which will not be important in determining the product series' sum expansion. So now the output of a(84) can be firmly stated to be {7, 7, 12, 14, 21, 28, 42, 84}.

Let's imagine that we are trying to determine the sum expansion of the product series when T=50. We will see the individual term 1/84 added 7 times (because there are 7 elements in a(84) that are below or equal to 50); here they are, how they'll appear when we FOIL the product series:

[itex]\frac{1}{2}*\frac{1}{42}*1*1*1...+\frac{1}{3}*\frac{1}{28}*1*1*1...+\fr ac{1}{4}*\frac{1}{21}*1*1*1...+\frac{1}{2}*\frac{1}{3}*\frac{1}{14}*1*1 *1...+\frac{1}{7}*\frac{1}{12}*1*1*1...+\frac{1}{3}*\frac{1}{4}*\frac{1 }{7}*1*1*1...+\frac{1}{2}*\frac{1}{6}*\frac{1}{7}*1*1*1...+[/itex](other elements, ≠1/84)

Do you see the relationship now?

EDIT: I give up trying to fix the itex tags. Hopefully you can still see what I'm getting at.

ramsey2879
#4
Dec3-11, 10:29 PM
P: 894
Certain and Curious Number Sequence

Quote Quote by Feryll View Post
EDIT: I just noticed something on reflection that, being me, I got wrong in the OP. If you end up with two or more repeats of output elements when figuring out the factorizations' greatest factors, you must repeat it that many times, which I didn't really specify. See below for implciations.

Thanks for the response, ramsey!

If you take that product series and examine how, up to any term, how you're going to manually multiply it out and what you're going to end up with, you should be able to see it. Look at this similar series:

[itex]∏_{n=2}^T (1+n^{-1})[/itex]
(1+1/2)(1+1/3)(1+1/4)...

Under these circumstances (the original power series /2), you add one (multiply out all ones, only), you add up the reciprocals all natural numbers up to and including T (pluck one n^-1 out, then all 1s), then all the reciprocals of the product of any two natural numbers up to and including T (pluck 2 n^-1s out, then all 1s), then all the reciprocals of the product of any three natural numbers up to and including T (pluck 3 n^-1s out, then all 1s), and so on, which is standard FOILing/associative law application.

And here you can connect it to the sequence.

Note that you will never be able to multiply out multiple identical n^-1s, so that is the reason that, in the original sequence, that if you end up with a factorization that contains two of the same factors, you must throw it out.

Now, for example, a(84) has the factorizations 2*2*3*7, 4*3*7, 2*6*7, 2*3*14, 2*2*21, 4*21, 7*12, 3*28, 2*42, 1*84. I've eliminated the illegal factorizations, as they have repeating factors, which will not be important in determining the product series' sum expansion. So now the output of a(84) can be firmly stated to be {7, 7, 12, 14, 21, 28, 42, 84}.

Let's imagine that we are trying to determine the sum expansion of the product series when T=50. We will see the individual term 1/84 added 7 times (because there are 7 elements in a(84) that are below or equal to 50); here they are, how they'll appear when we FOIL the product series:

[itex]\frac{1}{2}*\frac{1}{42}*1*1*1...+\frac{1}{3}*\frac{1}{28}*1*1*1...+\fr ac{1}{4}*\frac{1}{21}*1*1*1...+\frac{1}{2}*\frac{1}{3}*\frac{1}{14}*1*1 *1...+\frac{1}{7}*\frac{1}{12}*1*1*1...+\frac{1}{3}*\frac{1}{4}*\frac{1 }{7}*1*1*1...+\frac{1}{2}*\frac{1}{6}*\frac{1}{7}*1*1*1...+[/itex](other elements, ≠1/84)

Do you see the relationship now?

EDIT: I give up trying to fix the itex tags. Hopefully you can still see what I'm getting at.
I think I understand. T(8) = 9 = 2 + 2/2 + 2/3 + 2/4 + 2/5 + 4/6 + 2/7 + 4/8 + 2/10 + 4/12 + 2/14 + 2/15 + 2/16 + 2/18 + 2/20 + ... + 2/8!
Feryll
#5
Dec3-11, 10:42 PM
P: 7
Quote Quote by ramsey2879 View Post
I think I understand. T(8) = 9 = 2 + 2/2 + 2/3 + 2/4 + 2/5 + 4/6 + 2/7 + 4/8 + 2/10 + 4/12 + 2/14 + 2/15 + 2/16 + 2/18 + 2/20 + ... + 2/8!
Yep, that's it. (One of the things that depresses me about these problems is the difficulty in calculation; as you can see, we would have to evaluate all of the terms to over 40,000 just to calculate T(8); if someone could guide me to a simple algorithm to calculate the elements with a computer program I'd also be very thankful, as I'm weak with computing)

But nevertheless, I'm pretty certain this has been done before or is a magnification on some bigger problem previously explored, which is the work I'd like to see.


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