OFFSET
1,1
COMMENTS
Numbers k such that tau(k^2)/tau(k) = 5 where tau(n) is the number of divisors of n (A000005). - Bernard Schott, Nov 27 2020
LINKS
T. D. Noe, Table of n, a(n) for n = 1..1000
Will Nicholes, Prime Signatures
FORMULA
Sum_{n>=1} 1/a(n) = (P(2)^2*P(4) - P(4)^2)/2 - P(2)*P(6) + P(8) = 0.00125114..., where P is the prime zeta function. - Amiram Eldar, Jul 03 2022
a(n) = A085987(n)^2. - R. J. Mathar, May 05 2023
MATHEMATICA
f[n_]:=Sort[Last/@FactorInteger[n]]=={2, 2, 4}; Select[Range[200000], f]
PROG
(PARI) list(lim)=my(v=List(), t1, t2); forprime(p=2, (lim\36)^(1/4), t1=p^4; forprime(q=2, sqrt(lim\t1), if(p==q, next); t2=t1*q^2; forprime(r=q+1, sqrt(lim\t2), if(p==r, next); listput(v, t2*r^2)))); vecsort(Vec(v)) \\ Charles R Greathouse IV, Jul 24 2011
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Joseph Stephan Orlovsky, Jul 25 2010
STATUS
approved