20210314, 16:27  #980 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2×5×587 Posts 

20210314, 16:31  #981 
"Garambois JeanLuc"
Oct 2011
France
3×193 Posts 
OK, page updated.
Many thanks to all for your help. Added base : 58. New bases reserved for yoyo : 38, 43, 46 et 47. 74 bases in total. 
20210314, 16:33  #982 
"Garambois JeanLuc"
Oct 2011
France
3·193 Posts 

20210314, 22:23  #983  
Aug 2020
2×3^{2} Posts 
Quote:
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) == 3 (mod p) , so 3 must be quadratic residue mod p ] The smallest primes in that form are 11,13,23,37. So it will take many other small primes factor to make term at index 1 abundant. However, for each primes p_{i} == 1, 1 (mod 12), there exist odd k_{p} such that 3 ^ k_{i} == 1 (mod p) So we can choose as many primes p_{1}, p_{2},...p_{n} == 1, 1 (mod 12) as we need to make abundant term, taking product of all k_{i} of each prime p_{i}, then we will get 3 ^ (k_{1}*k_{2}*...*k_{n}) == 1 (mod p_{1}*p_{2}*...*p_{n}). k_{1}*k_{2}*...*k_{n} is odd, so there exist m such that 2*m+1= k_{1}*k_{2}*...*k_{n}. So s(3 ^ (2*m+1)) = (3 ^ (2m+2)  1) / 2 == (31)/2 == 1 (mod p_{1}*p_{2}*...*p_{n}), making the term at index 1 abundant. (this value m is likely very large.) 

20210314, 23:09  #984  
Aug 2020
2·3^{2} Posts 
Quote:


20210315, 18:16  #985  
"Garambois JeanLuc"
Oct 2011
France
3×193 Posts 
Quote:
Unfortunately, this exponent for base 30 is far too large for me to test with my program ! I'm assuming you got it by a similar method to what you expose in post #983 for base 3. Warachwe, after reading your last two posts, I think it is reasonable to stop my program for bases 3 and 30. The exponents which allow to obtain an abundant s(n) for these two bases are impossible to test with our current computers. I am going to test other bases and especially initially, the primorial bases after 6 and 30. I am going to test 210, 2310, 30030 ... Because it is all the same curious, that all the tested primorial bases b>210 have an abundant s(b^14), as the conjecture (134) says. I will point it out to you here if I notice any other curious things ... 

20210315, 22:10  #986 
Sep 2008
Kansas
6377_{8} Posts 
Starting work on base 62. This will be my last base of my choosing. We'll see what others have an interest in going forward.

20210315, 22:45  #987 
"Ed Hall"
Dec 2009
Adirondack Mtns
5×743 Posts 

20210316, 01:51  #988  
Aug 2020
2×3^{2} Posts 
Quote:
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. 

20210316, 04:51  #989 
"Curtis"
Feb 2005
Riverside, CA
2·3·797 Posts 
Factoring algorithms on generalform numbers such as these can be reasonably solved up to 180 digits or so, with 200 digits possible via concerted effort (and a few CPUyears of computation).
We can split off small factors up to 5060 digits fairly easily, so a number of roughly 240 digits has a reasonable chance of a full factorization (by finding small factors summing to 5070 digits, and cracking the rest with a full NFS algorithm). 
20210316, 16:52  #990 
"Garambois JeanLuc"
Oct 2011
France
3×193 Posts 
@RichD and EdH :
Thank you very much for the base 62. I will add it in the next update ... 
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