A224830 Numbers n such that both the sum of the semiprime divisors of n and the sum of the prime divisors of n are prime numbers.

36, 72, 108, 144, 165, 210, 216, 273, 288, 324, 345, 385, 399, 432, 462, 561, 576, 595, 648, 651, 665, 715, 795, 798, 858, 864, 885, 957, 972, 1001, 1015, 1110, 1152, 1218, 1281, 1290, 1296, 1335, 1443, 1463, 1495, 1515, 1533, 1547, 1551, 1615, 1645, 1659

Offset: 1

Author

Michel Lagneau, Jul 21 2013

Keywords

• nonn

Comments

Numbers n such that A008472(n) and A076290(n) are both prime numbers.

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.

Examples

			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.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Intersection of A114522 and A227680.

Cf. A008472, A076290.

Programs

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 *)