OFFSET
1,1
COMMENTS
There exists a subsequence of squares {36, 144, 324, 576, 1296, 2304, 2916, 5184, 9216, 11664, 20736, 26244, 36864, ...} and the numbers of the form n = (p*q)^2 or (p^a*q^v)^2 with p and q primes are in the sequence if we have the two conditions:
(1) p+q = p1 is prime => p=2
(2) p^2 + p*q + q^2 = p2 is prime (subsequence of A007645), because p^2, p*q and q^2 are the three possible semiprime divisors of n, but with p=2, the semiprime divisors are 4, 2q and q^2.
(1) and (2) => p2 - 2*p1 = q^2, hence the property:
Let a number n such that the sum of the semiprime divisors is a prime number p1 and the sum of the prime divisors of n is a prime number p2. If n is a perfect square having two prime divisors, then p1 - 2*p2 = 9. Proof:
If q > 3, q == 1 mod 6 => q^2 + 2q + 4 == 1 mod 6 (if q==5 mod 6, q^2 + 2q + 4 == 3 mod 6 is not prime), but q+2 == 3 mod 6 is not prime. Conclusion: q = 3, and q^2 = 9 if a(n) is a square.
Consequence: if a(n) is a square having two prime divisors, the number k*a(n) with k = 2 or 3 is in the sequence.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
EXAMPLE
72 is in the sequence because the sum of the prime divisors is 2+3 = 5 and the sum of the semiprime divisors is 4 + 2*3 + 9 = 19.
MAPLE
with(numtheory):for n from 2 to 2000 do:x:=divisors(n):n1:=nops(x): y:=factorset(n):n2:=nops(y):s1:=0:s2:=0:for i from 1 to n1 do: if bigomega(x[i])=2 then s1:=s1+x[i]:else fi:od: s2:=sum('y[i]', 'i'=1..n2):if type(s1, prime)=true and type(s2, prime)=true then printf(`%d, `, n):else fi:od:
MATHEMATICA
primeSum[n_] := Plus @@ First[Transpose[FactorInteger[n]]]; semipSigma[n_] := DivisorSum[n, # &, PrimeOmega[#] == 2 &]; Select[Range[2000], PrimeQ @ primeSum[#] && PrimeQ @ semipSigma[#] &] (* Amiram Eldar, May 10 2020 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Jul 21 2013
STATUS
approved