Number of sequences that merge with any given sequence - infinite?

[flagrantflowers]

Are the number of sequences that merger with any given sequence infinite in number? Does changing the definition of 'any give sequence' change the answer?

[science_man_88]

Originally Posted by flagrantflowers

Are the number of sequences that merger with any given sequence infinite in number? Does changing the definition of 'any give sequence' change the answer?

probably better threads for it but the number of sequences that hit any one given number n in a sequences is bounded on the upper side by the number of partitions with distinct parts of n. you can decrease that further to those that fit a aliquot divisor list. also it depends on what you call a merger all primes go to 1 so if you include the 1 in the sequence then all primes go to 1 and hence there are infinitely many there. see http://oeis.org/A048138

[Batalov]

Originally Posted by flagrantflowers

Are the number of sequences that merger with any given sequence infinite in number? Does changing the definition of 'any give sequence' change the answer?

Clearly not (unless you'd count merging at "1" a merger, but if you would the question would be almost meaningless although there are still cycles to concider).

Take sequence 2. Nothing merges into it and this is easy to prove.
Then, consider sequence 28. Again, nothing (except 28 itself).
Or sequence 3 (for which the single merging seq is alq(4)).
Etc.
__________________________

Forking this question now into a separate thread, because it has nothing to do with the subject of alq(4788).

[arbooker]

Originally Posted by flagrantflowers

Are the number of sequences that merger with any given sequence infinite in number?

Conditional on a strong form of the Goldbach conjecture, this is true for all odd numbers except 5, since you can find products of two primes that will merge with your number. For instance, writing \(s(n)=\sigma(n)-n\), \(3=s(4)=s(s(9))=s(s(s(15)))=s(s(s(s(33))))=\dots\).

For even numbers it's more complicated. For all of the ones that terminate, it's true by the above, since they will hit an odd prime different from 5. On the other hand, as Batalov pointed out, there are cycles consisting entirely of even numbers. Some of those also have infinitely many merging sequences (e.g. \(6=s(25)\) and \(284=s(80089)\)), but others provably do not (e.g. 28).

[science_man_88]

Originally Posted by arbooker

Conditional on a strong form of the Goldbach conjecture, this is true for all odd numbers except 5, since you can find products of two primes that will merge with your number. For instance, writing \(s(n)=\sigma(n)-n\), \(3=s(4)=s(s(9))=s(s(s(15)))=s(s(s(s(33))))=\dots\).

For even numbers it's more complicated. For all of the ones that terminate, it's true by the above, since they will hit an odd prime different from 5. On the other hand, as Batalov pointed out, there are cycles consisting entirely of even numbers. Some of those also have infinitely many merging sequences (e.g. \(6=s(25)\) and \(284=s(80089)\)), but others provably do not (e.g. 28).

problem is the sum of primes doesn't necessarily work for 7 as the sum of primes that equal 6 is 3 and 3 not distinct and hence not a divisors list. 2,4 leads to a divisors list for 7 of 1,2,4 leading back to 8.

edit:in fact given that argument you can say that until 2p for all prime p has at least 2 goldbach partitions you can't prove it. also if every odd number has to go back to a odd semiprime that implies either odd perfect numbers ( if they exist) must either be odd semiprimes ( I don't think their known form if they exist allows for that) or must have at least 2 sequences that merge with them directly.

[arbooker]

Originally Posted by science_man_88

problem is the sum of primes doesn't necessarily work for 7 as the sum of primes that equal 6 is 3 and 3 not distinct and hence not a divisors list. 2,4 leads to a divisors list for 7 of 1,2,4 leading back to 8.

It's likely true that every even number at least 8 is the sum of two distinct primes (and this is what I meant by a strong form of Goldbach's conjecture). The claim is still true for 7 since \(7=s(8)=s(s(49))=\dots\).

[science_man_88]

Originally Posted by arbooker

It's likely true that every even number at least 8 is the sum of two distinct primes (and this is what I meant by a strong form of Goldbach's conjecture). The claim is still true for 7 since \(7=s(8)=s(49)=\dots\).

now you have to guarantee it's not going to hit an untouchable number.

[arbooker]

Originally Posted by science_man_88

now you have to guarantee it's not going to hit an untouchable number.

Isn't that clear? Just keep going, e.g. \(49=s(215)=s(s(633))=\dots\). Each lift will be odd, so again we can find a product of two primes for the next one. Of course I can't prove that it will keep working since that would require proving the Goldbach conjecture, but that doesn't stop it being true.

[henryzz]

What are the largest untouchable numbers known?

[arbooker]

Originally Posted by henryzz

What are the largest untouchable numbers known?

It's known that a positive proportion of the natural numbers are untouchable, and in that sense there is no largest known. The proportion is conjectured to be about 17%, so they're fairly common.

I did a few experiments and was surprised to find that most aliquot cycles seem to be in the orbit of some odd number (and thus presumably infinitely many odd numbers). After 28, the next smallest counterexample is the amicable pair (356408,399592).

[science_man_88]

Originally Posted by arbooker

It's known that a positive proportion of the natural numbers are untouchable, and in that sense there is no largest known. The proportion is conjectured to be about 17%, so they're fairly common.

I did a few experiments and was surprised to find that most aliquot cycles seem to be in the orbit of some odd number (and thus presumably infinitely many odd numbers). After 28, the next smallest counterexample is the amicable pair (356408,399592).

but 2^74207281 -1 is the largest known prime without actually being the largest ... that's the thing about largest known it doesn't mean the same as largest.