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%I #40 Oct 31 2023 15:29:49
%S 3,5,3,34,53,3646,103,1367,57,5905,419,28483073,5317,2087804,12015,
%T 11658487,28913,551011827257,379419,987922247,249661,3328294201,
%U 1914791,9437402089436849,38755369,271566862001,4971027,7897940252308,351453749,1824613261627468511874644,233798623
%N a(n) is the least k such that n*k-1 has exactly n prime factors, counted with multiplicity.
%C If q is a prime that doesn't divide n, Dirichlet's theorem on primes in arithmetic progressions implies there are infinitely many k such that (k*n - 1)/q^(n-1) is a prime.
%H Robert Israel, <a href="/A366857/b366857.txt">Table of n, a(n) for n = 1..1049</a>
%F A001222(a(n)*n-1) = n.
%F a(n) >= (A053669(n)^n + 1) / n. - _David A. Corneth_, Oct 30 2023
%e a(4) = 34 because 4*34-1 = 135 = 3^3 * 5 has 4 prime factors, counted with multiplicity, and no smaller k works.
%p g:= proc(t,n) # first prime p == t mod n;
%p local p;
%p for p from t by n do if isprime(p) then return p fi od
%p end proc:
%p f:= proc(n) local m,t,p,P, T, Q, i, L, Lp,v,S; uses priqueue;
%p T:= select(i -> igcd(i, n)=1, [$1..n-1]);
%p P:= sort(map(g, T, n));
%p m:= nops(T);
%p initialize(Q); S:= {};
%p insert([-P[1]^n, [1$n]], Q);
%p do
%p t:= extract(Q);
%p if t[1] mod n = 1 then return (1-t[1])/n fi;
%p L:= t[2];
%p for i from 1 to n-1 do if L[i] < L[i+1] then
%p v:= t[1]*P[L[i]+1]/P[L[i]]; if member(v,S) then next fi;
%p S:= S union {v};
%p Lp:= subsop(i=L[i]+1, L); insert([v, Lp], Q)
%p fi od;
%p if L[n] < m then
%p v:= t[1]*P[L[n]+1]/P[L[n]];
%p if member(v,S) then next fi;
%p S:= S union {v};
%p Lp:= subsop(n=L[n]+1, L); insert([v, Lp], Q)
%p fi;
%p od;
%p end proc:
%p f(1):= 3:
%p map(f, [$1..40]);
%Y Cf. A001222, A053669.
%K nonn,look
%O 1,1
%A _Robert Israel_, Oct 30 2023
%E More terms from _David A. Corneth_, Oct 30 2023