Smooth power trios

[grandpascorpion]

That's great R. Coincedentally, I also found this answer yesterday (with the old program). Thanks ao much for your help with this idea. It was about 1.5 billion iterations into checking 59 6.

[R. Gerbicz]

At night running my smoothtrio program I've gotten:

17539913103192530=2*5*23*37*53*61*71*89*233*433
17539913103192531=3^6*17*149*223*277*367*419
17539913103192532=2^2*13^2*41*79*167*313*331*463

this is the smallest known smoothtrio for x=6
log(17539913103192532)/log(463)=6.09399 so this time not the strongest
solution.

So now grandpa you can reduce your computation to smoothtrio 54 6,
bound for primes=509 because this solution is <2^54. Now the expected
running time is only about 4 days!

[grandpascorpion]

Great stuff!

[Phil MjX]

Hi !
with a limit of 62 bits for d, the x=7 case has very little chance to succeed.
Is it a way to increase the d bound over 62 bits without having to rewrite all the program or loosing all the speed benefits?
Thanks.

[grandpascorpion]

New low solution found 3283630023973833,2,1. Running R's script now for 52 6

[R. Gerbicz]

Originally Posted by grandpascorpion

New low solution found 3283630023973833,2,1. Running R's script now for 52 6

Congratulations. ( I've stopped my program yesterday )
Here it is the factorizations:
3283630023973831=7^4*29*67*83*103*281*293
3283630023973832=2^3*13*19*43*59*97*107*223*283
3283630023973833=3^2*23*73*79*127*229*271*349

This is the smallest known solution for x=6.
The strength=log(3283630023973833)/log(349)=6.1020 ( not the strongest )

Originally Posted by Phil Mjx

Hi !
with a limit of 62 bits for d, the x=7 case has very little chance to succeed.
Is it a way to increase the d bound over 62 bits without having to rewrite all the program or loosing all the speed benefits?
Thanks.

It wouldn't be much slower, perhaps by a factor of two. I don't think that we have got a very little chance by smoothtrio 62 7. It would take less than 3 month for my computer.

[grandpascorpion]

The lowest answer is:
1348770149848002 = 2 x 3 x 7 x 23 x 41 x 61 ^ 2 x 149 x 239 x 257
1348770149848001 = 19 ^ 3 x 89 x 103 x 229 x 283 x 331
1348770149848000 = 2 ^ 6 x 5 ^ 3 x 11 x 29 x 109 x 151 x 163 x 197

log(1348770149848002)/log(2) = 50.26
log(1348770149848001)/log(331)=6.004353 (just barely a solution)

[R. Gerbicz]

That was a great work, garndpa!

If we use the strength of n is log(n)/log(q), where q is the largest prime factor of n. Then the smallest smoothtrio for x=2 is the following:
48=2^4*3
49=7^2
50=2*5^2
And the strength=log(49)/log(7)=2.
Grandpa this isn't an error in your first post, because you've used a slightly modified definition: smoothtrio's strength=log(n-2)/log(q), where q is the largest prime factor among n, n-1, n-2's prime factors.

[grandpascorpion]

Couldn't have done it without you, R!

Keep in mind, that in these x=6 cases, the differences between log(n)/log(p)
and log(n-1)/log(p) and log(n-2)/log(p) (where p is the high factor among the three numbers are almost negligible. For the last case, the values are the same through 15 decimal places. So, I'd rather keep the original definition even though I'm splitting hairs :)

BTW, could you give some insight into your last revision of this code? I'm really curious how you pared down the search. Thanks.


I suppose it's time for quartets now :)

[R. Gerbicz]

Originally Posted by grandpascorpion

BTW, could you give some insight into your last revision of this code? I'm really curious how you pared down the search. Thanks.

It isn't very hard, known algorithms. Some parts:
Generate all smooth numbers by lexicographic order
Let m=n*p, where p is the largest prime factor of m, then we want to see if m-1 is smooth or not. Let q be a prime(power) and suppose that m-1 is divisible by q, so
m-1==0 mod q, here m=n*p, so we can write:
n*p==1 mod q from this:
p==modinverse(n,q) mod q, so every q-th number is divisible by q. It means that by sieving we can save the trial factoring of m-1. I included your ideas also.

I've made a GMP version of the program. If you want I'll post. In that version: d<127 and d<=15*x are the limitations only.

[grandpascorpion]

R., that'd be great if you could post it. Thanks

Quartets (using a tweaked version of your code):

x=2

2106 = 2 x 3 ^ 4 x 13
2107 = 7 ^ 2 x 43
2108 = 2 ^ 2 x 17 x 31
2109 = 3 x 19 x 37
===================================
x=3

11503632 = 2 ^ 4 x 3 x 7 ^ 2 x 67 x 73
11503633 = 29 x 37 x 71 x 151
11503634 = 2 x 23 ^ 2 x 83 x 131
11503635 = 3 x 5 x 11 x 13 x 31 x 173
===================================
x=4

89861691840 = 2 ^ 6 x 3 x 5 x 23 x 73 x 197 x 283
89861691841 = 17 x 29 ^ 2 x 43 x 313 x 467
89861691842 = 2 x 13 x 151 x 191 x 293 x 409
89861691843 = 3 ^ 6 x 7 x 11 x 31 x 113 x 457

Strength: LOG(89861691841)/LOG(467) = 4.1035