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For 2-digit indices n, the following rules can be applied to expedite the search for a(n):
Let P(n) be the concatenation of a(n) and n. Then P(n) is the product of n primes (counted with multiplicity), P(n) mod 100 = n, a(n) = floor(P(n)/100) is prime, and it can be shown that the following constraints apply to the prime factors of P(n):
If n is odd, then 2 cannot appear among the prime factors of P(n).
If n == 2 (mod 4), then 2 must appear with multiplicity exactly 1.
If n == 4 (mod 8), then 2 must appear with multiplicity >= 3.
If n == 0 (mod 8), then 2 must appear with multiplicity exactly 2.
If n != 0 (mod 5), then 5 cannot appear among the prime factors of P(n).
If n == 0 (mod 5) but n != 0 (mod 25), then 5 must appear with multiplicity 1.
If n == 0 (mod 25), then 5 must appear with multiplicity >= 2.
Any prime q that divides n but does not divide 10 cannot appear among the prime factors of P(n).
For example, for n = 24, the following constraints apply to the primes that appear among the 24 prime factors of P(24):
since 8 | n, exactly two are 2's;
since 5 !| n, none are 5's;
since 3 | n, none are 3's;
so P(24) >= 2^2 * 7^22. As it turns out, P(24) = 123883192874635318124 = 2^2 * 7^19 * 11 * 13 * 19. (End)
a(924) = floor(2^2 * 13^907 * 17^9 * 19^5 * 31 / 1000) = 8.0881...*10^1026. - Jon E. Schoenfield, Mar 03 2023
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