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%I #33 Jun 22 2021 08:36:49
%S 6,20,42,78,118,184,248,332,428,534,654,772,906,1052,1208,1388,1562,
%T 1754,1958,2164,2396,2638,2896,3144,3424,3682,3986,4304,4622,4976,
%U 5286,5652,6002,6374,6748,7148,7532,7934,8356,8786,9224,9684,10158,10618,11114,11604
%N Smallest number k such that exactly half the numbers in [1..k] are prime(n)-smooth.
%C All terms are even numbers (because of the "exactly half the numbers in [1..k]" part of the definition).
%H Michael S. Branicky, <a href="/A290154/b290154.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..2000 from Robert Israel)
%H Michael S. Branicky, <a href="/A290154/a290154.txt">Python program for bfile</a>
%e 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.
%e 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.
%p N:= 100:
%p mypi:= proc(n) option remember; global pmax; local k;
%p k:= procname(pmax);
%p while pmax < n do pmax:= nextprime(pmax); k:= k+1 od;
%p k
%p end proc:
%p pmax:= 2: mypi(2):= 1:
%p V:= Vector(N):
%p count:= 0:
%p loheap:=heap[new](`<`,0): nlo:= 1:
%p hiheap:= heap[new](`>`,1): nhi:= 1:
%p for k from 4 by 2 while count < N do
%p for v in [mypi(max(numtheory:-factorset(k-1))), mypi(max(numtheory:-factorset(k)))] do
%p if v <= heap[max](loheap) then heap[insert](v,loheap); nlo:= nlo+1;
%p elif v >= heap[max](hiheap) then heap[insert](v,hiheap); nhi:= nhi+1;
%p elif nlo <= nhi then heap[insert](v,loheap); nlo:= nlo+1;
%p else heap[insert](v,hiheap); nhi:= nhi+1;
%p fi;
%p od;
%p if nlo < nhi-1 then
%p t:= heap[extract](hiheap);
%p heap[insert](t,loheap);
%p nlo:= nlo+1; nhi:= nhi-1;
%p elif nhi < nlo-1 then
%p t:= heap[extract](loheap);
%p heap[insert](t,hiheap);
%p nhi:= nhi+1; nlo:= nlo-1;
%p fi;
%p for n from heap[max](loheap) to min(heap[max](hiheap)-1, N) do
%p if V[n] = 0 then count:= count+1; V[n]:= k;
%p fi
%p od;
%p od:
%p convert(V,list); # _Robert Israel_, Mar 28 2019
%t 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 *)
%o (PARI) is(k,n) = {m=k; forprime(p=2, prime(n), while(m%p==0, m=m/p)); return(m==1); }
%o 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
%o (Python) # see link for a faster version producing bfile
%o from sympy import factorint, prevprime, primerange, prod
%o def aupto(limit):
%o adict, pN = dict(), prevprime(limit+1)
%o pi = {p: i for i, p in enumerate(primerange(1, pN+1), start=1)}
%o smooth = {i: 0 for i in pi.values()}
%o watching = smooth[0] = 1 # 1 is prime(n) smooth for all n
%o for n in range(2, limit+1):
%o f = factorint(n, limit=pN)
%o nt = prod(p**f[p] for p in f if p <= pN)
%o if nt == n: smooth[pi[max(f)]] += 1
%o if 2*sum(smooth[i] for i in range(watching+1)) == n:
%o adict[watching] = n
%o watching += 1
%o return sorted(adict.values())
%o print(aupto(12000)) # _Michael S. Branicky_, Jun 20 2021
%Y Cf. A000079, A003586, A126283.
%K nonn
%O 1,1
%A _Jon E. Schoenfield_, Jul 21 2017
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