three: click to enlarge |

five: click to enlarge |

seven: click to enlarge |

manual distribution worksheet for 59536 primality tests |

What have I discovered? Of Leyland numbers with at least one million decimal digits, but fewer than one million one hundred decimal digits, there is only one prime. That prime was discovered by Gabor Levai long before I got to it. I saved all of my primality-test output where the 59536 entries are listed smallest to largest. If you want to see the one prime, search for "PRP".

The execution times vary wildly (due to processor circumstances) with an average of 9.45 hours per test. That would work out to 64 years if I hadn't been able to multi-process. I know now how to keep the execution times to 6 hours or less per test but that means running fewer processes per machine. Still, I might be able to shave a couple of months off the total time required for the next run.

Éric Angelini presented a proposal for three "digit-spine" sequences on his blog here, as well as to the MathFun community. I decided to take on the first one:

s = 1, 10, 2, 0, 3, 26, 9, 119, 532, 4, 6, 896, 118, 34, 15, ...

p = 2, 11, 2, 2, 3, 23, 7, 113, 523, 3, 5, 887, 113, 31, 13, ...

d = 1, 1, 0, 2, 0, 3, 2, 6, 9, 1, 1, 9, 5, 3, 2, ...

Our sought-after sequence is 's'. This is followed by 'p', a sequence of examples (for each, there might be a second such) of primes that are as close to the terms of 's' as possible. Finally 'd', the absolute differences between the respective 's' and 'p' values. Determining 's' is that it must be the lexicographically earliest sequence of distinct nonnegative integers such that its digit-spine (the sequential digits) as well as the digit-spine of 'd' are identical.

For example, the third 'd' being 0 is a consequence of the zero in 10, the second term of 's'. This forces the third term of 's' to be the smallest prime (as it had not already been used). And so on. There will always be integers in 'd' that inform our upcoming values in 's'. And the digits of those values get added to the growing end of our 'd' sequence.

So far, all of our 'd' values are single digits. But there will come a point when concatenating a digit with the next one in line allows for a smaller 's' term than what is available for the single digit. This first happens at term #1622 where a d = 1 dictates s = 1010. The next d digit is 0. Concatenating that 0 with the previous 1 gives us d = 10 at #1622 and this allows the smaller s = 897. This makes me want to know at which term the first three-digit d appears.

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