Good mathematical tinkerers are in the Zen room

[R.D. Silverman]

Originally Posted by LaurV

As it was clear from the context what "mostly composite" means, see your post #6. I just auctioned on it. As usual, we see the straw in other people's eye, but can't see the wooden beam in our...

Moron.

No, it was NOT clear. Maybe if you ever take the trouble to learn some math, you will understand it.

"mostly composite" can have a number of meanings. One does NOT use colloquial English when
discussing math. One DOES use proper quantifiers. There is a big difference between pointing out
the improper use of a quantifier, and (your) claiming my lack of rigor because the domain of a variable was not
explicitly stated but rather derived from the context of the problem.

[science_man_88]

Originally Posted by science_man_88

I'll be waiting to hear from you with your answers.

I should add that this is not the first number of this sort but an absolute minimum without distinct prime divisor lower limits involved. edit: which means you can prove a k-multiperfect number can't be less than 8*k-10

[wildrabbitt]

5) is it not possible that if the formula for the n-th triangular number includes dividing by 2 that the numerator in the formula must be 2*(99*m) = 198*m for the number to be a 100-multiperfect number ?

That's wrong.

the nth triangular was given as a maximal limit on sigma(n). sigma(n) could be less than this limit /* EDIT : and still 100 times n */ and therefore whether or not

the numerator of the the (m/2)th triangular number is 2*99*m is irrelevant.

[science_man_88]

Originally Posted by wildrabbitt

That's wrong.

the nth triangular was given as a maximal limit on sigma(n). sigma(n) could be less than this limit and therefore whether or not

the numerator of the the (m/2)th triangular number is 2*99*m is irrelevant.

it was given as a maximal sum for the n=m/2 for proper divisors only which is not sigma(n) but sigma(n)-n which from previous statements and logic we can prove for a k-multiperfect number m is (k-1)*m we then can show that if that value takes triangular form that the numerator is effectively 2*(k-1)*m =1/4* m^2 +(m/2) dividing both sides by m we get 2*(k-1) = 1/4*m-1/2 ; taxing both sides as halves after the addition to cancel out 1/2 we get: (4*(k-1)-1)/2 = (1/2*m)/2 multiplying both by 2 we get (4*(k-1)-1) = (1/2*m) dividing by 1/2 ( aka multiplying by 2) we get 8*(k-1)-2 = m or 8*k-10 =m as a minimum possible.

[wildrabbitt]

Thanks for explaining that. I thought I'd found a fault.
I don't get it all yet but I'll keep trying to understand it.

[wildrabbitt]

4) is it taking the rest into account, possible to say with certainty that if the (m/2)th triangular number is less than 99 *m that m can't be a 100-multiperfect number ?

The answer to this question being YES does not mean that 100-perfect numbers can't exist :

You made an implication :

if the (m/2)th triangular number is less than 99 *m that m can't be a 100-multiperfect number ?

This implication is true so you can say it truly but you haven't proved the hypothesis which is

the (m/2)th triangular number is less than 99 *m

You haven't proved A so you can't claim B is true which is :

m can't be a 100-multiperfect number

[science_man_88]

Originally Posted by wildrabbitt

Thanks for explaining that. I thought I'd found a fault.
I don't get it all yet but I'll keep trying to understand it.

maybe an example is in order:

lets try to find the smallest number that could possibly be a 2-multiperfect number m ( another name for perfect numbers from that link I posted) :

lets assume my answer for the minimum is correct with k=2, we get 8*2-10 = m = 16-10 =6;

we know that m divides by 1 that can be mostly ignored:
1/2/3)we then go on and say what is the maximum number when divided by a number greater than 1 that m can be divisible by and we get m/2 lets call this n, so we know that the proper divisors ( divisors that are not the number itself) can range from 1 to m/2, lets say for simplicity that we sum these numbers to get a maximum above which the sum of proper divisors can't pass. this sum will be T(m/2) =T(n) we know for the number to be 2-multiperfect that the sum of all divisors has to be 2*m subtracting the number itself like we do for a sum of proper divisors leaves us with (2-1)*m we know the form of T(n) is (n*(n+1))/2.

4) so as a sum of proper divisors of a 2-multiperfect number m, we get (2-1)*m =< T(n) we know that otherwise our maximum sum of proper divisors T(n) can't make it to 2*m when m is added

5) assume they equal for a second and our 2-multiperfect number testing can work out. so 1*m=T(n)=(m/2)*((m/2)+1)/2 with n replaced by the m/2 we originally assigned it. multiplying by 2 to get rid of that denominator we get:

6) 2*1*m=2*T(n) = (m/2)*((m/2)+1) =(m^2)/(2^2) +m/2 =1/4*m^2+m/2 dividing out m we get:

7) 2=1/4*m+1/2 so we can put everything in halves we continue by multiplying the other numerators by 2 keeping their values the same when divided by 2:

8/9/10) 4/2 = (1/2*m)/2 +1/2 which goes to 3/2 = (1/2*m)/2 multiply both by two and we get 3=1/2*m dividing both by half gives 6=m just like our prediction said it would. in this case 6 is the lowest the others aren't always going to make the cut as actual k-multiperfect numbers though.

[science_man_88]

Originally Posted by wildrabbitt

The answer to this question being YES does not mean that 100-perfect numbers can't exist :

You made an implication :

if the (m/2)th triangular number is less than 99 *m that m can't be a 100-multiperfect number ?

This implication is true so you can say it truly but you haven't proved the hypothesis which is

the (m/2)th triangular number is less than 99 *m

You haven't proved A so you can't claim B is true which is :

m can't be a 100-multiperfect number

I'm not saying they don't exist what I'm saying is if that maximal sum of proper divisors isn't at very least 99*m then m can't be a 100-multiperfect number because the sum of all the divisors wouldn't hold.

[wildrabbitt]

You've said

statement A : the m/2th triangular number is less than 99m
statement B : there cannot be any 100-perfect numbers

and

A implies B therefore B

which is assuming without proof that A is true. Until you've proved A you've not proved B.

There are many m for which 99m is less than the m/2th triangular number.

If you're whole argument rests on this mistake it's wrong.

[science_man_88]

Originally Posted by wildrabbitt

You've said

statement A : the m/2th triangular number is less than 99m
statement B : there cannot be any 100-perfect numbers

and

A implies B therefore B

which is assuming without proof that A is true. Until you've proved A you've not proved B.

There are many m for which 99m is less than the m/2th triangular number.

If you're whole argument rests on this mistake it's wrong.

I was actually doing the statement about they must be large because technically until you get into distinct prime divisors minimums 790 is the lowest possible number with minimal properties needed for it to be a 100-multiperfect number. I didn't say anything about there not being any, what I did say is if the (m/2)th triangular number is less than 99 *m that m can't be a 100-multiperfect number which is true because if the sum of proper divisors isn't greater than or equal to 99*m then the sum of divisors including m can't reach 100*m T(m/2) is the maximum this sum of divisors could be so if it's not greater than or equal to 99*m there's no way m is a 100-multiperfect number.

[wildrabbitt]

So what is anybody supposed to conclude when they've got to step 10 apart from 800-10 = 790 ?