[QUOTE=garambois;573695]Have you had enough or do you still need more ?
Should I run the program even further ?[/QUOTE]

You have found enough.

OK, page updated.
Many thanks to all for your help.

[B]Added base :[/B] 58.
[B]New bases reserved for yoyo :[/B] 38, 43, 46 et 47.

74 bases in total.

[QUOTE=henryzz;573696]You have found enough.[/QUOTE]

OK, many thanks...

[QUOTE=garambois;573610]
This is my new holy grail : a base 3 sequence with an odd exponent whose term at index 1 would be abundant ![/QUOTE]

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) == 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[SUB]i[/SUB] == 1, -1 (mod 12), there exist odd k[SUB]p[/SUB] such that 3 ^ k[SUB]i[/SUB] == 1 (mod p)
So we can choose as many primes p[SUB]1[/SUB], p[SUB]2[/SUB],...p[SUB]n[/SUB] == 1, -1 (mod 12) as we need to make abundant term, taking product of all k[SUB]i[/SUB] of each prime p[SUB]i[/SUB],
then we will get 3 ^ (k[SUB]1[/SUB]*k[SUB]2[/SUB]*...*k[SUB]n[/SUB]) == 1 (mod p[SUB]1[/SUB]*p[SUB]2[/SUB]*...*p[SUB]n[/SUB]).

k[SUB]1[/SUB]*k[SUB]2[/SUB]*...*k[SUB]n[/SUB] is odd, so there exist m such that 2*m+1= k[SUB]1[/SUB]*k[SUB]2[/SUB]*...*k[SUB]n[/SUB].

So s(3 ^ (2*m+1)) = (3 ^ (2m+2) - 1) / 2 == (3-1)/2 == 1 (mod p[SUB]1[/SUB]*p[SUB]2[/SUB]*...*p[SUB]n[/SUB]), making the term at index 1 abundant.
(this value m is likely very large.)

[QUOTE=garambois;573694]I tried my program with the base 30.
Nothing at all after 24 hours of calculation !!!
While it only took a few seconds for bases 6, 12 and 24 ...

I will be doing some other tests in the next few days and try to figure out which are the even and odd bases for which it is very difficult to find sequences (with base and exponent of the same parity) that have abundant index 1 terms ![/QUOTE]

If I am doing caculation correctly, s(30^(12*(23#))) should be abundant.

[QUOTE=warachwe;573724]If I am doing caculation correctly, s(30^(12*(23#))) should be abundant.[/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 ...

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.

[QUOTE=RichD;573808]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.[/QUOTE]I see one merge already:
[code]
Running base 62 from 1 through 100 . . .
62^11:i260 merges with 38664:i4
[/code]

[QUOTE=garambois;573791]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.
[/QUOTE]

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.

[QUOTE=warachwe;573819]How high can you test with base 30?
[/QUOTE]
Factoring algorithms on general-form numbers such as these can be reasonably solved up to 180 digits or so, with 200 digits possible via concerted effort (and a few CPU-years of computation).

We can split off small factors up to 50-60 digits fairly easily, so a number of roughly 240 digits has a reasonable chance of a full factorization (by finding small factors summing to 50-70 digits, and cracking the rest with a full NFS algorithm).

@RichD and EdH :
Thank you very much for the base 62.
I will add it in the next update ...