Forum: Aliquot Sequences
2021-08-09, 16:13
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Replies: 1,818
Views: 211,569
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Forum: Aliquot Sequences
2021-04-05, 01:33
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Replies: 1,818
Views: 211,569
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...
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Forum: Aliquot Sequences
2021-03-24, 19:53
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Replies: 1,818
Views: 211,569
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Forum: Aliquot Sequences
2021-03-24, 07:10
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Replies: 1,818
Views: 211,569
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 *...
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Forum: Aliquot Sequences
2021-03-22, 20:44
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Replies: 1,818
Views: 211,569
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.
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Forum: Aliquot Sequences
2021-03-16, 19:23
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Replies: 1,818
Views: 211,569
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...
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Forum: Aliquot Sequences
2021-03-16, 01:51
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Replies: 1,818
Views: 211,569
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Forum: Aliquot Sequences
2021-03-14, 23:09
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Replies: 1,818
Views: 211,569
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Forum: Aliquot Sequences
2021-03-14, 22:23
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Replies: 1,818
Views: 211,569
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) ==...
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Forum: Aliquot Sequences
2021-03-06, 18:19
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Replies: 1,818
Views: 211,569
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 *...
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Forum: Aliquot Sequences
2021-03-04, 15:45
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Replies: 1,818
Views: 211,569
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.
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Forum: FactorDB
2020-11-20, 01:47
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Replies: 71
Views: 19,028
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...
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Forum: FactorDB
2020-11-15, 21:16
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Replies: 71
Views: 19,028
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...
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Forum: Aliquot Sequences
2020-09-14, 09:44
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Replies: 1,818
Views: 211,569
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Forum: Aliquot Sequences
2020-08-22, 11:03
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Replies: 1,818
Views: 211,569
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) =...
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Forum: Aliquot Sequences
2020-08-20, 18:56
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Replies: 1,818
Views: 211,569
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...
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Forum: Aliquot Sequences
2020-08-20, 18:30
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Replies: 1,818
Views: 211,569
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...
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Forum: Aliquot Sequences
2020-08-20, 03:45
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Replies: 1,818
Views: 211,569
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Forum: Aliquot Sequences
2020-08-19, 16:25
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Replies: 1,818
Views: 211,569
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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