Set of particular numbers

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Goahead52
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Set of particular numbers

Postby Goahead52 » Sun Oct 04, 2015 11:26 am UTC

Hi,

n > 0

n=abcd.....z where a,b,c,...z are digits

A=a^a+b^b+c^c.....+Z^z

By convention 0^0=1

Find all the couples (n,A) such as gdc(n,A)=1
The first :
(11,2),(12,5),(13,(13,28),(14,257)....

Is the set of such numbers infinite?
What are the main properties of such couples?

Thank you for your thoughts.

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mathmannix
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Re: Set of particular numbers

Postby mathmannix » Tue Oct 06, 2015 6:03 pm UTC

If n is the product of 26 "digits" (aka integers 0-9) AND n>0 then :

(1) n is the product of 26 integers 1-9, i.e. none of them are zero. (So your statement 0^0=1 is irrelevant.) But that would mean that A would be at least 26 (if a...z were all 1), so it couldn't be 2 or 5 as in your examples.

(2) the set is definitely not infinite, because if the 26 numbers a...z are each 1 through 9, then there are "only" a total of 9^26 (approximately 6.46 septillion) possible combinations to check.
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Sizik
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Re: Set of particular numbers

Postby Sizik » Tue Oct 06, 2015 6:19 pm UTC

I think they mean n = a*10^25 + b*10^24 + c*10^23 + ... + z * 10^0. So for the first example (n = 11), a thorough x are 0, and y & z are 1.

Edit:
Unless they also mean that n can be an arbitrary integer, and A is the sum of each digit of n to the power of itself.
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Goahead52
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Re: Set of particular numbers

Postby Goahead52 » Tue Oct 06, 2015 7:16 pm UTC

There is a big misunderstanding here.
When I noted a,b,c....z it does not mean 26 digits.
It could be 100, 1000, and finite number.
Sorry for that.

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Re: Set of particular numbers

Postby Goahead52 » Tue Oct 06, 2015 7:20 pm UTC

n=1542161348148184911354411541152255412252411152234115511124111221111

A=1^1+5^5+4^4+2^2+...and so on until + 1^1+1^1+1^1

From the first digit to the last one.

n can have any number of digits from 1 to the "infinite" (finite number).

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Re: Set of particular numbers

Postby ConMan » Tue Oct 06, 2015 10:29 pm UTC

I'm not sure about the properties of these pairs, but I suspect that it has the same issue as any set of numbers defined by something to do with their digits, which is that it's highly dependent on the base chosen, and 10 is not actually that interesting a base to work in.

However, if you switch to binary and look at the same kind of property, then that quickly becomes looking at the set of numbers whose binary digit sum is coprime to them, and that might be of some interest, somehow.
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Goahead52
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Re: Set of particular numbers

Postby Goahead52 » Tue Oct 06, 2015 11:27 pm UTC

@Conman

I agree with your comments.
My goal in fact is to partition the set of integers in 2 or more sets such as if we compute some summatory functions of arithmetical functions (moebius, Euler totient etc...) we will find some properties. No one can know what are the interesting properties of a set before computations. A lot of work is required before reaching that goal.
For now I have to compute at least the first 100.000 elements of the set. I`m doing it on Excel (I`m not a programmer).
It takes time.

Thank you very much.

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Re: Set of particular numbers

Postby Goahead52 » Wed Oct 07, 2015 10:31 am UTC

Let aside the question of partitioning the set of integers N.
If we define A=f(n)

Let us start with n=2
f(2)=2^2-=4
f*4)=4^4=256
f(256)=2^2+5^5+6^6=49785
f(49785)...... we continue iterating the same function...
What will happen?
Are we going at some step to come back to some cycle or it is going to diverge?
Does the sequence generated will be always composite or no?
Where it will lead us?

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Re: Set of particular numbers

Postby Xanthir » Wed Oct 07, 2015 5:55 pm UTC

Well, 9^9 is only ~400 million, so the maximum value of f() for an N-digit number is about N*400 million. This means that roundabout a dozen digits (I don't wanna math right now), f(a) < a for all a. So yeah, you're guaranteed to cycle.

I highly doubt you're guaranteed to stay composite.
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Re: Set of particular numbers

Postby curiosityspoon » Thu Oct 08, 2015 9:02 am UTC

Starting from n=2, the sequence eventually settles into a loop of 40 repeating numbers, which first stabilizes with the 20th term: 17653671, 1743552, 830081, 33554462, 53476, 873607, 18470986, 421845378, 34381644, 16824695, 404294403, 387421546, 17651084, 17650799, 776537847, 20121452, 3396, 387467199, 793312220, 388244100, 33554978, 405027808, 34381363, 16824237, 17647707, 3341086, 16824184, 33601606, 140025, 3388, 33554486, 16830688, 50424989, 791621836, 405114593, 387427281, 35201810, 16780376, 18517643, 17650825. None of these numbers are prime, although there are a few semiprimes including 34381363 = 1051 x 32713. Starting from 3, 4, 6, or 7 gives the same result.

Other starting points such as 5, 8, or 9 produce a different loop of 97 numbers; in the case of n=9, this loop even starts as early as the third term in the sequence. Furthermore, the 97-cycle does include three prime numbers: 405067961, 59159, and 826673.

Among all starting points up to n=1000, every sequence so far has eventually settled into one of 5 cycles: the trivial 1-cycle of n=1, the aforementioned 40- and 97-cycles, an 11-cycle (17604196, 388337603, 34424740, 824599, 791621579, 776537851, 19300779, 776488094, 422669176, 388384265, 50381743) first encountered at n=188, and a 3-cycle (16777500, 2520413, 3418) first encountered at n=345. There will also be a 1-cycle at 3435 (first encountered at 3345). Obviously, if a given starting point ends up in a particular loop, every digit permutation of that number will yield the same sequence after the first term, so without loss of generality you can assume the digits in the starting number are sorted in ascending order.

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Re: Set of particular numbers

Postby Goahead52 » Thu Oct 08, 2015 2:01 pm UTC

@curiosityspoon

Thank you for all those details.
It is easy to know the cycle of each number element of the cycle but how to find the root become hard problem at first glance.
How can we find 2 as root and before 2 the 10 and 11 starting from some element (for example 17653671)?
2 is out of the loop. I mean the starting point is not included in the loop.
For some number like 3435 where f(3435)=3435 the number is a part of the loop. Are there an infinite of numbers such as fn)=n?
Anyway those cycles are fascinating and need more research I think.
I`m wondering if some odd semi-prime numbers share one factor of the elements of their cycle. If there are a lot then it will lead to easy factorization (at least for those numbers). I do not think so it seems highly unlikely.
Lot of questions.
Are those sequences related to any other problem in number theory?

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Re: Set of particular numbers

Postby PM 2Ring » Sat Oct 10, 2015 6:20 am UTC

Goahead52 wrote:Are those sequences related to any other problem in number theory?


Probably not. As Conman said earlier, sequences like this don't usually reveal anything particularly deep about the numbers involved because the results depend on the base you're using to represent the numbers. There's nothing of particular mathematical significance in using ten as a base. The only (integer) base that can be considered special is binary, and sometimes you can get interesting properties from these "numerological" sequences when you use 2 as the base (as Conman also said).

So you can't just look at a bunch of sequences starting with various seed numbers, you have to look at what happens under a variety of different bases, too. You might get similar interesting-looking sequences happening in every base, or you might only get the interesting-looking behaviour in some bases; if you can find find some way of predicting which bases make the nice sequences then you may have found an interesting property.

FWIW, a classic example of a base-dependent property that has been investigated extensively is the cyclic numbers.

Goahead52
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Re: Set of particular numbers

Postby Goahead52 » Sat Oct 10, 2015 2:37 pm UTC

@PM2ring

Thanks for your comments.
I agree with you but for someone like me who is not programmer it is not easy to try other bases (2,3,6,12,60,...).
So i started from base 10 which is is a common base.
So if n is prime > 7 there is only one prime p=157 such as p divide f(p). It is strange and needs some explanation.
My goal was to slit the set of all positive integers into finite number of sets (here 4 sets).
Why?
I wanted to isolate some particular set (either composed by only composite numbers either by square-free number either by unique cycle etc...)
I have introduced a new function based on digits.
I still did not study it until because it is little complicated to compute on Excel.

Goahead52
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Re: Set of particular numbers

Postby Goahead52 » Sat Oct 10, 2015 2:37 pm UTC

@PM2ring

Thanks for your comments.
I agree with you but for someone like me who is not programmer it is not easy to try other bases (2,3,6,12,60,...).
So i started from base 10 which is is a common base.
So if n is prime > 7 there is only one prime p=157 such as p divide f(p). It is strange and needs some explanation.
My goal was to slit the set of all positive integers into finite number of sets (here 4 sets).
Why?
I wanted to isolate some particular set (either composed by only composite numbers either by square-free number either by unique cycle etc...)
I have introduced a new function based on digits.
I still did not study it until because it is little complicated to compute on Excel.

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Re: Set of particular numbers

Postby PsiSquared » Mon Oct 12, 2015 9:26 pm UTC

Goahead52 wrote:@curiosityspoon
For some number like 3435 where f(3435)=3435 the number is a part of the loop. Are there an infinite of numbers such as fn)=n?


Obviously not.

As Xanthir said, for n>1011 f(n) will always be less than n. So for all n>1011, the equation f(n)=n would be false.

In reality, 1 and 3435 are the only numbers for which f(n)=n.

As for your original question:
Find all the couples (n,A) such as gdc(n,A)=1

Is the set of such numbers infinite?


I don't know how to find ALL such couples, but the set is clearly infinite:

Let p be any prime greater than 3 and let n=1111...1111 with p 1's.

Clealry in this case A=p. so gcd(n,A)=1 if and only if p doesn't divide A.

But according to Fermat little theorem, p divides 1111...1111 with (p-1) 1's. Let us call this number X.

Now observe that n=10X+1.

X is divisible by p, so clearly n = 1 (mod p).

In other words: p does not divide n.

Which means that gcd(n,A)=1

And since p can be any prime you wish (as long as it is greater than 3), I've proven that there is an infinite number of pairs such that gcd(n,A)=1.


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