Fun with Numbers: The Pan-Prime-Digit Cube Hypothesis

[Werner]

Well, that is the challenge as numbers keep growing ;-)
For x < 1e17, I can cube it exactly quite easily, as Free42 carries 34 digits, so
x^3 calculated as x.x^2 has a single roundoff error

Then,
x^3 = B + r, where
r = MOD(x^3,1e17) = MOD(x.MOD(x^2,1e17),1e17) (for x<1e17, x^2 is exact)
r are the 17 trailing digits of x^3
B = x^3 - r, a number (< 1e51) ending in 17 zeroes.

By tomorrow I will have passed 1e51 for x^3 and I'll have to find something new.
(I already did btw)

PS. Currently running with the 20 million numbers < 1e12 that, when cubed, have the trailing 12 digits in {2,3,5,7} and narrowing down the cubes to match the first 8 digits. Verification of a number focuses first on the digits 'in the middle' as we know that the 8 leading and 12 trailing digits are all 2,3,5 and 7
Verifying cubes larger than 1e50 and smaller than 1e51 will take 3.5 hours, estimated.
Pre-computing an extra order of numbers speeds up the process by a factor 2.5, but order 12 seems to be the maximum, memory-wise. REGS can hold the 20 million, but not the 80 million needed for order 13.

Cheers, Werner