20210314, 16:27  #980 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
3×1,973 Posts 

20210314, 16:31  #981 
"Garambois JeanLuc"
Oct 2011
France
7×97 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
7×97 Posts 

20210314, 22:23  #983  
Aug 2020
19_{10} 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
19 Posts 
Quote:


20210315, 18:16  #985  
"Garambois JeanLuc"
Oct 2011
France
7·97 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
2^{3}·431 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
4001_{10} Posts 

20210316, 01:51  #988  
Aug 2020
19 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
3^{3}·5·37 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
2A7_{16} Posts 
@RichD and EdH :
Thank you very much for the base 62. I will add it in the next update ... 
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