OT: an original conjecture by my alter-ego.

[Rudy Toody]

Edit: there might be a counter-example, so everything is sitting for a while.
This is Rudy Toody tooting his own horn! The conjecture.

 

[petrusbroder]

I wish I understood more ...
Happy New Year, Fred!

 

[Rudy Toody]

I wish I understood more ...
Happy New Year, Fred!

Back at you, Peter.

Mathematicians claim that anyone can come up with a conjecture. The difficulty is coming up with an interesting conjecture. I've had comments that this is a nice problem.

 

[Ken g6]

It would be nice if you could spell it out a little more in plain English.

If I'm reading it right...

B(n,k) is the largest number d <= k where d is a factor of n. I.e. The largest not-necessarily-prime factor of n <= k.

S(n,k) is the sum of those B(n,k) for a given k. I.e. if the Sieve of Eratosthenes is run dumbly (with both primes and composites), over a range n=1...k! with maximum prime factor k, this is the sum of the largest factors that hit each element? (I notice that the n passed in is not used.)

And T(n,k) is the number of (distinct?) products of all numbers in {1..k} and {1..k!}? (Again, I notice that the n passed in is not used.)

Edit:

Happy New Year, Fred!

This too! It seems like it's too early (in the year) to be dealing with such heavy math.

 

[Rudy Toody]

It would be nice if you could spell it out a little more in plain English.

If I'm reading it right...

B(n,k) is the largest number d <= k where d is a factor of n. I.e. The largest not-necessarily-prime factor of n <= k.

S(n,k) is the sum of those B(n,k) for a given k. I.e. if the Sieve of Eratosthenes is run dumbly (with both primes and composites), over a range n=1...k! with maximum prime factor k, this is the sum of the largest factors that hit each element? (I notice that the n passed in is not used.)

And T(n,k) is the number of (distinct?) products of all numbers in {1..k} and {1..k!}? (Again, I notice that the n passed in is not used.)

Edit:
This too! It seems like it's too early (in the year) to be dealing with such heavy math.

Thanks, Ken for the explanation. It's right on the mark.

We're calling them divisors instead of factors, because factors generally mean primes. And the products are distinct because # as set cardinality implies no duplicates.

I didn't notice the n not being used. I'll have to fix that. Edit---My co-conspirator says, "Nice catch."

 

[Ken g6]

Hm, now it's down to functions of one variable, this should be easy to check for a few steps. The OEISes you linked don't match each other, but they don't seem to match your functions exactly either.

 

[Rudy Toody]

Hm, now it's down to functions of one variable, this should be easy to check for a few steps. The OEISes you linked don't match each other, but they don't seem to match your functions exactly either.

Keep trying. They should match the first OEIS for k!

The second OEIS is for a sub-symmetric sequence. I guess I should add a comment that the formula is found on OEIS. (I had both versions displayed, but it was suggested that I eliminate one since they are very similar---only the limits change.)