A290154 Smallest number k such that exactly half the numbers in [1..k] are prime(n)-smooth.

6, 20, 42, 78, 118, 184, 248, 332, 428, 534, 654, 772, 906, 1052, 1208, 1388, 1562, 1754, 1958, 2164, 2396, 2638, 2896, 3144, 3424, 3682, 3986, 4304, 4622, 4976, 5286, 5652, 6002, 6374, 6748, 7148, 7532, 7934, 8356, 8786, 9224, 9684, 10158, 10618, 11114, 11604

Offset: 1

Author

Jon E. Schoenfield, Jul 21 2017

Keywords

• nonn

Comments

All terms are even numbers (because of the "exactly half the numbers in [1..k]" part of the definition).

Examples

			The 2-smooth numbers are 1, 2, 4, 8, 16, 32, ... (A000079, the powers of 2), so the numbers of 2-smooth numbers in the interval [1..k] for k = 2, 4, and 6 are 2, 3, and 3, respectively; thus, the smallest k at which the number of 2-smooth numbers in [1..k] is exactly k/2 is k=6, so a(1)=6.
The 3-smooth numbers are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, ... (A003586), so there are more than k/2 3-smooth numbers in [1..k] for every positive k < 20, but exactly k/2 3-smooth numbers in [1..20], so a(2) = 20.
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000 (terms 1..2000 from Robert Israel)
• Michael S. Branicky, Python program for bfile

Crossrefs

Cf. A000079, A003586, A126283.

Programs

Maple

N:= 100:
mypi:= proc(n) option remember; global pmax; local k;
  k:= procname(pmax);
  while pmax < n do pmax:= nextprime(pmax); k:= k+1 od;
  k
end proc:
pmax:= 2: mypi(2):= 1:
V:= Vector(N):
count:= 0:
loheap:=heap[new](`<`,0): nlo:= 1:
hiheap:= heap[new](`>`,1): nhi:= 1:
for k from 4 by 2 while count < N do
   for v in [mypi(max(numtheory:-factorset(k-1))), mypi(max(numtheory:-factorset(k)))] do
     if v <= heap[max](loheap) then heap[insert](v,loheap); nlo:= nlo+1;
     elif v >= heap[max](hiheap) then heap[insert](v,hiheap); nhi:= nhi+1;
     elif nlo <= nhi then heap[insert](v,loheap); nlo:= nlo+1;
     else heap[insert](v,hiheap); nhi:= nhi+1;
     fi;
   od;
   if nlo < nhi-1 then
        t:= heap[extract](hiheap);
        heap[insert](t,loheap);
        nlo:= nlo+1; nhi:= nhi-1;
   elif nhi < nlo-1 then
        t:= heap[extract](loheap);
        heap[insert](t,hiheap);
        nhi:= nhi+1; nlo:= nlo-1;
   fi;
   for n from heap[max](loheap) to min(heap[max](hiheap)-1, N) do
     if V[n] = 0 then count:= count+1; V[n]:= k;
     fi
   od;
od:
convert(V,list); # Robert Israel, Mar 28 2019

Mathematica

smoothQ[k_, p_] := k <= p || Max[FactorInteger[k][[All, 1]]] <= p; a[n_] := For[p = Prime[n]; cnt = 0; k = 1, True, k++, If[smoothQ[k, p], cnt++]; If[cnt == k/2, Return[k]]]; Array[a, 46] (* Jean-François Alcover, Jul 22 2017 *)

PARI

is(k,n) = {m=k; forprime(p=2, prime(n), while(m%p==0, m=m/p)); return(m==1); }
a(n) = {j=2; x=2; y=0; while(x!=y, j+=2; s=is(j,n)+is(j-1,n); x+=s; y+=2-s); j; } \\ Jinyuan Wang, Aug 03 2019

Python

# see link for a faster version producing bfile
from sympy import factorint, prevprime, primerange, prod
def aupto(limit):
    adict, pN = dict(), prevprime(limit+1)
    pi = {p: i for i, p in enumerate(primerange(1, pN+1), start=1)}
    smooth = {i: 0 for i in pi.values()}
    watching = smooth[0] = 1  # 1 is prime(n) smooth for all n
    for n in range(2, limit+1):
        f = factorint(n, limit=pN)
        nt = prod(p**f[p] for p in f if p <= pN)
        if nt == n: smooth[pi[max(f)]] += 1
        if 2*sum(smooth[i] for i in range(watching+1)) == n:
            adict[watching] = n
            watching += 1
    return sorted(adict.values())
print(aupto(12000)) # Michael S. Branicky, Jun 20 2021