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

3, 2, 83, 2, 67, 41, 947, 4519, 15659081, 2843, 337957, 389, 1616171, 6132829, 422116888343, 24850181, 377519743, 194486417892947, 533348873, 324403, 980825013273164555563, 25691144027, 273933405157, 1238831928746353181, 311195507789, 129917586781, 2159120477658983490299

Offset: 1

Author

Zak Seidov and Robert Israel, Feb 24 2023

Keywords

• nonn
• base

Comments

From Jon E. Schoenfield, Feb 26 2023: (Start)

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

Examples

			a(5) = 67 because 67 is prime and 675 = 3^3 * 5^2 with A001222(675) = 3+2 = 5, and no smaller prime works.
		

Links

• Jon E. Schoenfield, Table of n, a(n) for n = 1..923

Crossrefs

Cf. A001222, A066686.

Programs

Maple

icat:= proc(a,b) 10^(1+ilog10(b))*a+b end proc:
f:= proc(n) local p;
p:= 1;
do
  p:= nextprime(p);
  if numtheory:-bigomega(icat(p,n)) = n then return p fi;
od
end proc:
map(f, [$1..17]); # Robert Israel, Feb 24 2023

Formula

A001222(A066686(a(n),n)) = n.

Extensions

a(18) and a(21)-a(27) from Jon E. Schoenfield, Feb 24 2023