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A179669
Products of form p^4*q^2*r where p, q and r are three distinct primes.
9
720, 1008, 1200, 1584, 1620, 1872, 2268, 2352, 2448, 2736, 2800, 3312, 3564, 3920, 4050, 4176, 4212, 4400, 4464, 5200, 5328, 5508, 5808, 5904, 6156, 6192, 6768, 6800, 7452, 7500, 7600, 7632, 7938, 8112, 8496, 8624, 8784, 9200, 9396, 9648, 9680, 10044
OFFSET
1,1
MAPLE
N:= 20000: # for terms < N
P:= select(isprime, [2, seq(i, i=1..N/(3^2*2^4), 2)]):
R:= NULL:
for i from 1 while P[i]^4 * 2^2*3 < N do
for j from 1 while P[i]^4 * P[j]^2 *2 < N do
if j = i then next fi;
m:= ListTools:-BinaryPlace(P, N/P[i]^4/P[j]^2);
R:= R, seq(P[i]^4*P[j]^2*P[k], k={$1..m} minus {i, j});
od od:
sort([R]); # Robert Israel, Mar 28 2025
MATHEMATICA
f[n_]:=Sort[Last/@FactorInteger[n]]=={1, 2, 4}; Select[Range[10000], f]
PROG
(PARI) list(lim)=my(v=List(), t1, t2); forprime(p=2, (lim\12)^(1/4), t1=p^4; forprime(q=2, sqrt(lim\t1), if(p==q, next); t2=t1*q^2; forprime(r=2, lim\t2, if(p==r||q==r, next); listput(v, t2*r)))); vecsort(Vec(v)) \\ Charles R Greathouse IV, Jul 20 2011
(Python)
from math import isqrt
from sympy import primepi, primerange, integer_nthroot
def A179669(n):
def bisection(f, kmin=0, kmax=1):
while f(kmax) > kmax: kmax <<= 1
kmin = kmax >> 1
while kmax-kmin > 1:
kmid = kmax+kmin>>1
if f(kmid) <= kmid:
kmax = kmid
else:
kmin = kmid
return kmax
def f(x): return n+x-sum(primepi(x//(p**4*q**2)) for p in primerange(integer_nthroot(x, 4)[0]+1) for q in primerange(isqrt(x//p**4)+1))+sum(primepi(integer_nthroot(x//p**4, 3)[0]) for p in primerange(integer_nthroot(x, 4)[0]+1))+sum(primepi(isqrt(x//p**5)) for p in primerange(integer_nthroot(x, 5)[0]+1))+sum(primepi(x//p**6) for p in primerange(integer_nthroot(x, 6)[0]+1))-(primepi(integer_nthroot(x, 7)[0])<<1)
return bisection(f, n, n) # Chai Wah Wu, Mar 28 2025
CROSSREFS
Cf. A137493.
Sequence in context: A337036 A257416 A137493 * A067808 A302127 A291804
KEYWORD
nonn
AUTHOR
STATUS
approved