Smallest 10^179+c Brilliant Number (p90 * p90)

[Dr Sardonicus]

Originally Posted by Alfred

Dr Sardonicus,

does this statement apply to largest 2-brilliant numbers in base 10?

If yes, please give an example.

No. The argument I had in mind only applies to the smallest 2-brilliant number N > b^2k, an even power of the base b > 1. The argument is a triviality. If

N = p*q, p < q prime

with b^2k < N < b^2k+1

then, in order that p and q have the same number of base-b digits, it is necessary that p >= b^k. (The only possibility of equality is with k = 1 when the base b is prime.)

The only question that arises is whether the indicated number p₁p₂ is less than b^2k+1. I believe this will be true for sufficiently large k.

I see no analogous argument for least brilliant greater than an odd power of the base, or for greatest brilliant number less than any power of the base.

[Alfred]

Dr Sardonicus,

thank you for your explanation.

[Branger]

Continuing this work for 10^169-c, I have found that,

10^169-14319 =
2093963760229909907466815025292144577767961972509185032132596865267781491968551925027 *
4775631837535734107517020048684519409802862518997809812035307071144108182496827007803

I also attach proof files for some of the work I have done, for 10^n+-c for n = 167 and 169. Every number that has a factor larger than 1000 has that factor listed in the files. I'll post the files for n=165 shortly, I seem to have lost some ECM work that I'll redo first.

I intend to continue with n=171, but now I'm starting to get into the territory where the SNFS polynomials are getting rather large coefficients for the batch factorization approach and the relations I already have saved. I'm not sure if it would be quicker to sieve again for a new shared rational side, or if using the bad polynomials with the already existing relations is the least work, but for now I'm using what I have.

[Alfred]

I think it is a good idea to share these informations.

Thank you.

So anyone who is interested in can doublecheck the

correctness of the statements easily.

[Branger]

And finally here are the proof files for 10^165+-c.

[swishzzz]

I am reserving 10^199+c to find the smallest 200-digit number which splits into p100*p100. Likely to take at least a few months with an expected 160+ SNFS factorizations, thought I'd at least post here to prevent any potential duplicated efforts. If anyone is interested in crunching a few of these let me know, I can coordinate sieving efforts on another thread.

[Branger]

The next one was quicker and only required 27 SNFS factorizations.

10^171+7467 =

15982339170654488061693029140006521400812407348641102533477071444640746972955602480993 *
62569063847432371483112919249240694575724386807642240564815844097979821472243476190219

Continuing with 10^171-c.

[Branger]

Another 60 SNFS factorizations revealed that

10^171-16569 =
10026073074372053022855343749617316836566548448825765741868691467529514155418582501607 *
99739947293634841017115301301296053264053136150611286806831026018776115664482913987233

Proof file is attached. The batch SNFS approach was still faster than regular SNFS but its getting close, I'll probably continue with 10^173 +- c