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by madcaptenor 460 days ago
10^10 * 36105/39062500 = 9242880, so you're already down to under 10^7 cases to check, which is starting to seem more tractable.
1 comments

10^7 cases, but almost every case has billions of digits.

Even that doesn't seem so bad though, it's on the order of 10^16 total digits to check in the worst case, and far fewer in practice.

Maybe someone here can run a program overnight and increase the bound by another few orders of magnitude, or disprove the hypothesis?

You only need to check enough digits to find an odd one. An odd digit appears in the low order (< 46 up to a high level) digits for the first quadrillion cases so you only need to compute 2^n mod 10^d where d is big enough to be safe. I used d=60 in my computations to take this to 10^15 candidates (with no additional terms found).