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A189991
Numbers with prime factorization p^4*q^4.
6
1296, 10000, 38416, 50625, 194481, 234256, 456976, 1185921, 1336336, 1500625, 2085136, 2313441, 4477456, 6765201, 9150625, 10556001, 11316496, 14776336, 17850625, 22667121, 29986576, 35153041, 45212176, 52200625, 54700816, 57289761, 68574961, 74805201
OFFSET
1,1
COMMENTS
The primes p and q must be distinct, or else the product has factorization p^8 (or q^8, for that matter).
FORMULA
Sum_{n>=1} 1/a(n) = (P(4)^2 - P(8))/2 = (A085964^2 - A085968)/2 = 0.000933..., where P is the prime zeta function. - Amiram Eldar, Jul 06 2020
MATHEMATICA
lst = {}; Do[If[Sort[Last/@FactorInteger[n]] == {4, 4}, Print[n]; AppendTo[lst, n]], {n, 55000000}]; lst (* Orlovsky *)
lim = 10^8; pMax = PrimePi[(lim/16)^(1/4)]; Select[Union[Flatten[Table[Prime[i]^4 Prime[j]^4, {i, 2, pMax}, {j, i - 1}]]], # <= lim &] (* Alonso del Arte, May 03 2011 *)
With[{nn=30}, Take[Union[Times@@@(Subsets[Prime[Range[nn]], {2}]^4)], nn]] (* Harvey P. Dale, Mar 05 2015 *)
PROG
(PARI) list(lim)=my(v=List(), t); forprime(p=2, lim^(1/8), t=p^4; forprime(q=p+1, (lim\t)^(1/4), listput(v, t*q^4))); vecsort(Vec(v)) \\ Charles R Greathouse IV, Jul 24 2011
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved