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Forum: Aliquot Sequences 2022-09-14, 16:59
Replies: 1,957
Views: 223,577
Posted By warachwe
I think, with no evidence to back it up, it might...

I think, with no evidence to back it up, it might be because 43 have many values of "reverse aliquot" (don't know what it's actually called) in the first few terms. So it's more like grouping...
Forum: Aliquot Sequences 2021-08-09, 16:13
Replies: 1,957
Views: 223,577
Posted By warachwe
The problem is that when the factor 13 are with...

The problem is that when the factor 13 are with even power, it does not preserve the 7 for the second iteration.
This only happen when k is multiple of 13, for example 2^(12*13)-1...
Forum: Aliquot Sequences 2021-04-05, 01:33
Replies: 1,957
Views: 223,577
Posted By warachwe
The exponent "i" chosen this way is design to...

The exponent "i" chosen this way is design to make many small primes p divide s(b^i), but this is only work for p dividing neither the base b, nor p divide (q-1) for any of prime q in the factor of...
Forum: Aliquot Sequences 2021-03-24, 19:53
Replies: 1,957
Views: 223,577
Posted By warachwe
I use the same method that @kruoli point out, and...

I use the same method that @kruoli point out, and thank you very much for checking those factors.
Forum: Aliquot Sequences 2021-03-24, 07:10
Replies: 1,957
Views: 223,577
Posted By warachwe
3^5 * 19^2 * 29^2 * 31 * 37^2 * 47 * 59 * 83 *...

3^5 * 19^2 * 29^2 * 31 * 37^2 * 47 * 59 * 83 * 103 * 109 * 131 * 139 * 199 * 223 * 271 * 307 * 419 * 523 * 613 * 647 * 691 * 701 * 757 * 811 * 859 * 1063 * 1093 * 1123 * 1151 * 1231 * 1259 * 1381 *...
Forum: Aliquot Sequences 2021-03-22, 20:44
Replies: 1,957
Views: 223,577
Posted By warachwe
I tried a few number of exponents on base 7. ...

I tried a few number of exponents on base 7.
So far 1987441237556775=3^5 · 5^2 · 7 · 11 · 13 · 17 · 19 · 23 · 29 · 37 · 41 is lowest odd n I found that s(7^n) is abundant, but with primes <105.
Forum: Aliquot Sequences 2021-03-16, 19:23
Replies: 1,957
Views: 223,577
Posted By warachwe
Thank you for the information. However, as...

Thank you for the information. However, as @garambois said, we don't need whole factorisation.


@garambois The 12 * (23 #) is just an exponent that I'm quite sure would be abundant, but very...
Forum: Aliquot Sequences 2021-03-16, 01:51
Replies: 1,957
Views: 223,577
Posted By warachwe
How high can you test with base 30? It might...

How high can you test with base 30?
It might work for exponent 2^4*3^2*5*7*11=55440, or even some lower exponents (15120, 27720,30240, etc).
If not, 2^4*3^3*5*7*11=166320 should work.
Forum: Aliquot Sequences 2021-03-14, 23:09
Replies: 1,957
Views: 223,577
Posted By warachwe
If I am doing caculation correctly,...

If I am doing caculation correctly, s(30^(12*(23#))) should be abundant.
Forum: Aliquot Sequences 2021-03-14, 22:23
Replies: 1,957
Views: 223,577
Posted By warachwe
I think this is going to be hard to find. If a...

I think this is going to be hard to find.
If a prime p divide s(3 ^ (2k+1)), then p must be 1 or -1 (mod 12)

[ for odd prime p, s(3 ^ (2k+1)) =(3 ^ (2k+2) - 1) / 2 == 0 (mod p) iff 3 ^ (2k+2) ==...
Forum: Aliquot Sequences 2021-03-06, 18:19
Replies: 1,957
Views: 223,577
Posted By warachwe
For p prime, and and a such that gcd(a,p)=1, we...

For p prime, and and a such that gcd(a,p)=1, we must have a^(p-1) == 1 mod p.
So if p-1 divides 2^3 * 3^2 * 5 * 7, it means 2^(2^3 * 3^2 * 5 * 7 * k) == 1^k == 1 mod p, hence 2^(2^3 * 3^2 * 5 * 7 *...
Forum: Aliquot Sequences 2021-03-04, 15:45
Replies: 1,957
Views: 223,577
Posted By warachwe
38^(12*19) has next term...

38^(12*19) has next term (http://factordb.com/sequences.php?se=1&aq=38%5E%2812*19%29&action=range&fr=0&to=20) as 3*5*7*13*229*457*C353. As the composite has no small factor, this is not abundant.
Forum: FactorDB 2020-11-20, 01:47
Replies: 73
Views: 19,910
Posted By warachwe
This is what I guess was happened. On 15th...

This is what I guess was happened.
On 15th November, someone enter a number of form bn with b<20000 and n<1000 or somewhere around that ballpark. Since then DB slowly determined whether each number...
Forum: FactorDB 2020-11-15, 21:16
Replies: 73
Views: 19,910
Posted By warachwe
I notice that most of the new number is of the...

I notice that most of the new number is of the form a^n+-1 for 10001<=a<=20000 and n somewhere around 20. Most of these numbers already factored at cownoise.com . Is there anyway to just transfered...
Forum: Aliquot Sequences 2020-09-14, 09:44
Replies: 1,957
Views: 223,577
Posted By warachwe
The problem seem to affect only n that is divided...

The problem seem to affect only n that is divided by some prime bigger than 50. Somehow it change 72^n to 2^(6*n).
Forum: Aliquot Sequences 2020-08-22, 11:03
Replies: 1,957
Views: 223,577
Posted By warachwe
Personally, I think these conjectures are...

Personally, I think these conjectures are interesting in a way that the prove seem somewhat doable, but I agree that we should look more into other phenomena.



I think I found a way.
S(24k) =...
Forum: Aliquot Sequences 2020-08-20, 18:56
Replies: 1,957
Views: 223,577
Posted By warachwe
From post #364 for p prime, s(pi) =...

From post #364


for p prime, s(pi) = (pi-1)/(p-1) and s(p(i*n)) = (p(i*n)-1)/(p-1).

Since (pi-1) is a factor of (p(i*n)-1), s(pi) is a factor of s(p(i*n)) for all positive integer n.

for...
Forum: Aliquot Sequences 2020-08-20, 18:30
Replies: 1,957
Views: 223,577
Posted By warachwe
Sure :smile: Those conjectures may...

Sure :smile:



Those conjectures may still be true, because we can still get a factor of 79 from other primes than 157.
For example, from conjecture 2), normally 5 is what provide a factor of 3...
Forum: Aliquot Sequences 2020-08-20, 03:45
Replies: 1,957
Views: 223,577
Posted By warachwe
I forgot to mention that q^2 also must not divide...

I forgot to mention that q^2 also must not divide n, or else p may not divide s(n).



So in this case, for s(s(s(n))) may not be divided by 79 if, for example, s(s(n)) got a factor of 157^2, but...
Forum: Aliquot Sequences 2020-08-19, 16:25
Replies: 1,957
Views: 223,577
Posted By warachwe
For p prime, s(p^k) is (p^k-1)/(p-1). So...

For p prime, s(p^k) is (p^k-1)/(p-1). So S(3^(6+12*k)) is (3^(6+12*k)-1)/2, which is (3^6-1)/2 * (3^(12*k)+3^(12k-6)+3^(12k-12)...+1)

So s(n) divided by (3^6-1)/2 = 364 = 2^2 *7 * 13.

As 3^6 is...
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