Fun with Numbers: The PanPrimeDigit Cube Hypothesis

08102017, 08:17 PM
Post: #32




RE: Fun with Numbers: The PanPrimeDigit Cube Hypothesis
(08092017 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 order11 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 order11 numbers as brute force doesn't work with a range of 100 billion numbers ;) I use numbers A + b, where b is the 11digit precomputed part and A = a.10^11. Then (A + b)^3 = (Xr) + 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 Xr 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 

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