A106413 Smallest number beginning with 3 that is the product of exactly n distinct primes.
3, 33, 30, 330, 3570, 30030, 3015870, 30120090, 300690390, 30043474230, 304075581810, 30035662366710, 304250263527210, 30078810535603830, 3001252188252588270, 32589158477190044730, 3003056284355533696290
Offset: 1
Examples
a(1) = 3, a(6) = 30030 = 2*3*5*7*11*13.
Links
- Chai Wah Wu, Table of n, a(n) for n = 1..45
Programs
-
Maple
f:= proc(n) uses priqueue; local pq, t, p, x, i,L,v,Lp; initialize(pq); L:= [seq(ithprime(i),i=1..n)]; v:= convert(L,`*`); insert([-v, L], pq); do t:= extract(pq); x:= -t[1]; if floor(x/10^ilog10(x)) = 3 then return x fi; L:= t[2]; p:= nextprime(L[-1]); for i from n to 1 by -1 do if i < n and L[i] <> prevprime(L[i+1]) then break fi; Lp:= [op(L[1..i-1]),op(L[i+1..n]),p]; insert([-convert(Lp,`*`),Lp], pq) od od; end proc: map(f, [$1..30]); # Robert Israel, Sep 12 2024
-
Python
from itertools import count from math import prod, isqrt from sympy import primerange, integer_nthroot, primepi, primorial def A106413(n): if n == 1: return 3 def g(x,a,b,c,m): yield from (((d,) for d in enumerate(primerange(b+1,isqrt(x//c)+1),a+1)) if m==2 else (((a2,b2),)+d for a2,b2 in enumerate(primerange(b+1,integer_nthroot(x//c,m)[0]+1),a+1) for d in g(x,a2,b2,c*b2,m-1))) def f(x): return int(sum(primepi(x//prod(c[1] for c in a))-a[-1][0] for a in g(x,0,1,1,n))) for l in count(len(str(primorial(n)))-1): kmin, kmax = 3*10**l-1, 4*10**l-1 mmin, mmax = f(kmin), f(kmax) if mmax>mmin: while kmax-kmin > 1: kmid = kmax+kmin>>1 mmid = f(kmid) if mmid > mmin: kmax, mmax = kmid, mmid else: kmin, mmin = kmid, mmid return kmax # Chai Wah Wu, Sep 12 2024