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2021-03-06, 16:18   #953
henryzz
Just call me Henry

"David"
Sep 2007
Liverpool (GMT/BST)

135508 Posts

Quote:
 Originally Posted by garambois This is proved for all k integers such that 0 < k < 5001. It remains to prove it for all integer values ​​of k, but this is too difficult for me. Indeed, s(6^(2^3 * 3^2 * 5 * 7)) = d * c, with d = 5^2 * 7^2 * 11 * 13 * 19 * 29 * 31 * 37 * 41 * 43 * 61 * 71 * 73 * 127 * 181 * 211 * 281 * 337 * 421 * 631. But, d is abundant, because s(d) / d > 1.2. And using a very small program, it is very easy to verify for all k from 1 to 5000 that d is a divisor of s(6^((2^3 * 3^2 * 5 * 7) * k)). On the other hand, I don't know how to prove that d divides the terms of index 1 for all k.
I think the trick would be to prove that:
2^(2^3 * 3^2 * 5 * 7 * k + 1) - 1 == 1 mod d
3^(2^3 * 3^2 * 5 * 7 * k + 1) - 1) / 2 == 1 mod d
6^(2^3 * 3^2 * 5 * 7 * k) == 1 mod d

If this is true then s(6^(2^3 * 3^2 * 5 * 7 * k)) = (2^(2^3 * 3^2 * 5 * 7 * k + 1) - 1)*3^(2^3 * 3^2 * 5 * 7 * k + 1) - 1) / 2 - 6^(2^3 * 3^2 * 5 * 7 * k) = 1*1-1 mod d

Quote:
 Conjecture (139) : There exists for each base b a starting exponent i, such that for any integer k, the sequences b^(i * k) are increasing from index 1 during at least one iteration. One last little idea To finish, I would like to present to you an idea which should allow us to demonstrate that the conjecture for the base 38 was false, with the i = 3 * 5 * 7 * 13 which is deficient. I wrote a little program. I calculate for as many k as possible the numbers c = 38^(12 * k) / i = 38^(12 * k) / (3 * 5 * 7 * 13). And then I do a primality test on c. If c is a prime number for a single value of k, the conjecture is invalidated.
Surely you mean c = s(38^(12 * k)) / i

I suspect that the way to attack this for individual b at least would be to find an i that has a d that is abundant.

As far as I can tell it is usually quite easy. Finding that for 22^(180 * k), d= 3^3*7*13*19*31*5*37*61 which is abundant was quite easy.

d=3^2*5*7*13 is abundant.
2^(180 * k + 1) - 1 == 1 mod d
p^(180 * k) == 1 mod d for a lot of p. I haven't worked out the logic behind this yet. http://factordb.com/index.php?query=...at=1&sent=Show
(2*p)^(180 * k) == 1 mod d if gcd(d,p)==1 as indicated by http://factordb.com/index.php?query=...at=1&sent=Show

I would be very interested if someone could come up with a case where (2*p)^(180*k) does not have an abundant divisor for some p.