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A358596 a(n) is the least prime p such that the concatenation p|n has exactly n prime factors with multiplicity.

%I A358596 #24 Mar 03 2023 22:35:02
%S A358596 3,2,83,2,67,41,947,4519,15659081,2843,337957,389,1616171,6132829,
%T A358596 422116888343,24850181,377519743,194486417892947,533348873,324403,
%U A358596 980825013273164555563,25691144027,273933405157,1238831928746353181,311195507789,129917586781,2159120477658983490299
%N A358596 a(n) is the least prime p such that the concatenation p|n has exactly n prime factors with multiplicity.
%C A358596 From _Jon E. Schoenfield_, Feb 26 2023: (Start)
%C A358596 For 2-digit indices n, the following rules can be applied to expedite the search for a(n):
%C A358596 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):
%C A358596 If n is odd, then 2 cannot appear among the prime factors of P(n).
%C A358596 If n == 2 (mod 4), then 2 must appear with multiplicity exactly 1.
%C A358596 If n == 4 (mod 8), then 2 must appear with multiplicity >= 3.
%C A358596 If n == 0 (mod 8), then 2 must appear with multiplicity exactly 2.
%C A358596 If n != 0 (mod 5), then 5 cannot appear among the prime factors of P(n).
%C A358596 If n == 0 (mod 5) but n != 0 (mod 25), then 5 must appear with multiplicity 1.
%C A358596 If n == 0 (mod 25), then 5 must appear with multiplicity >= 2.
%C A358596 Any prime q that divides n but does not divide 10 cannot appear among the prime factors of P(n).
%C A358596 For example, for n = 24, the following constraints apply to the primes that appear among the 24 prime factors of P(24):
%C A358596   since 8 | n, exactly two are 2's;
%C A358596   since 5 !| n, none are 5's;
%C A358596   since 3 | n, none are 3's;
%C A358596 so P(24) >= 2^2 * 7^22. As it turns out, P(24) = 123883192874635318124 = 2^2 * 7^19 * 11 * 13 * 19. (End)
%C A358596 a(924) = floor(2^2 * 13^907 * 17^9 * 19^5 * 31 / 1000) = 8.0881...*10^1026. - _Jon E. Schoenfield_, Mar 03 2023
%H A358596 Jon E. Schoenfield, <a href="/A358596/b358596.txt">Table of n, a(n) for n = 1..923</a>
%F A358596 A001222(A066686(a(n),n)) = n.
%e A358596 a(5) = 67 because 67 is prime and 675 = 3^3 * 5^2 with A001222(675) = 3+2 = 5, and no smaller prime works.
%p A358596 icat:= proc(a,b) 10^(1+ilog10(b))*a+b end proc:
%p A358596 f:= proc(n) local p;
%p A358596 p:= 1;
%p A358596 do
%p A358596   p:= nextprime(p);
%p A358596   if numtheory:-bigomega(icat(p,n)) = n then return p fi;
%p A358596 od
%p A358596 end proc:
%p A358596 map(f, [$1..17]); # _Robert Israel_, Feb 24 2023
%Y A358596 Cf. A001222, A066686.
%K A358596 nonn,base
%O A358596 1,1
%A A358596 _Zak Seidov_ and _Robert Israel_, Feb 24 2023
%E A358596 a(18) and a(21)-a(27) from _Jon E. Schoenfield_, Feb 24 2023