Post Reply 
Fun with Numbers: The Pan-Prime-Digit Cube Hypothesis
08-10-2017, 08:17 PM
Post: #32
RE: Fun with Numbers: The Pan-Prime-Digit Cube Hypothesis
(08-09-2017 03:20 AM)DavidM Wrote:  Care to share some pseudocode for your algorithm?

I change it every day ;-)

I'm currently running with all 5'010'048 order-11 numbers precomputed, that is, all
numbers < 1e11, that, when cubed, have their 11 least significant digits in {2,3,5,7}.
This is the largest set I can use, as the next order will exceed 1e34 when cubed.
I had to write a specific routine to compute the 5 million order-11 numbers as brute force doesn't work with a range of 100 billion numbers ;-)

I use numbers A + b, where b is the 11-digit precomputed part and A = a.10^11.
Then (A + b)^3 = (X-r) + r

X = (A+b)^3, rounded to 34 digits
r = MOD(b^3,1e11), numbers known to consist only of 2,3,5 and 7
X-r is a number that ends in 11 zeroes, as the exact value of (A+b)^3 ends with the 11 digits of r.
This way I can go up to 1e34x1e11 = 1e45.

Then, I started narrowing down the numbers to verify, as follows:
All cubed numbers must start with either 2,3,5 or 7. For each of these, they must even be between x.222...eYY and x.777...eYY.
So I loop over

loop
(2.222.. eyy)^(1/3) < A < (2.777.. eyy)^(1/3)
(3.222.. eyy)^(1/3) < A < (3.777.. eyy)^(1/3)
(5.222.. eyy)^(1/3) < A < (5.777.. eyy)^(1/3)
(7.222.. eyy)^(1/3) < A < (7.777.. eyy)^(1/3)
yy := yy+1;
end loop;

A is increased in steps of 1e11, and at each step all 5 million numbers A+b are verified. Each step takes about 3 minutes ;-). At that rate I'll reach A=1e15 in 5 days. Trouble is, David Hayden has already gone that far and found nothing, and I'm not sure how to go beyond ;-)

Werner

41CV†,42S,48GX,49G,DM42,DM41X,17BII,15CE,DM15L,12C,16CE
Find all posts by this user
Quote this message in a reply
Post Reply 


Messages In This Thread
RE: Fun with Numbers: The Pan-Prime-Digit Cube Hypothesis - Werner - 08-10-2017 08:17 PM



User(s) browsing this thread: 2 Guest(s)