A345452 Positive integers with an even number of prime factors (counting repetitions) that sum to an even number.

1, 4, 9, 15, 16, 21, 25, 33, 35, 36, 39, 49, 51, 55, 57, 60, 64, 65, 69, 77, 81, 84, 85, 87, 91, 93, 95, 100, 111, 115, 119, 121, 123, 129, 132, 133, 135, 140, 141, 143, 144, 145, 155, 156, 159, 161, 169, 177, 183, 185, 187, 189, 196, 201, 203, 204, 205, 209, 213, 215

Offset: 1

Author

Peter Munn, Jun 20 2021

Keywords

• nonn
• easy

Comments

Numbers with an even number of even prime factors and an even number of odd prime factors.

The representation (as defined in A206284) of polynomials with nonnegative integer coefficients that are in the ideal of the polynomial ring Z[x] generated by x^2+x and 2.

The above property arises because the sequence lists the integers in the multiplicative subgroup of positive rational numbers generated by the squares of primes (A001248) and the products of two consecutive odd primes (A006094\{6}).

The sequence is closed under multiplication, prime shift (A003961), and - where the result is an integer - under division. Using these closures, all the terms can be derived from the presence of 4 and 15. For example, A003961(4) = 9, A003961(9) = 25, A003961(15) = 35, 15 * 35 = 525, 525/25 = 21. Alternatively, the sequence may be defined as the closure of A046337 under multiplication by 4.

From the properties of subgroups of the positive rationals we know that if we take an absent positive integer m and divide all terms that are multiples of m by m, we get all the integers in the same subgroup coset as m, and we can expect some of the nice properties here to carry over to the resulting set. Specifically, dividing the even terms by 2 gives all numbers with an odd number of prime factors that sum to an even number; dividing all terms divisible by an odd prime p by p, gives all numbers with an odd number of prime factors that sum to an odd number. The positive integers satisfying the 4th of the 4 possibilities are generated similarly, dividing by 6 (for example).

Numbers whose squarefree part is in A056913.

Term by term, the sequence is one half of its complement within A036349.

Examples

			The definition specifies that we count repeated prime factors.
6 = 2 * 3; the sum of these prime factors is 2 + 3 = 5, an odd number; so 6 is not in the sequence.
50 = 2 * 5 * 5 has 3 prime factors and 3 is an odd number; so 50 is not in the sequence.
60 = 2 * 2 * 3 * 5 has 4 prime factors and 4 is an even number; the sum of these factors is 2 + 2 + 3 + 5 = 12, also an even number; so 60 is in the sequence.
1 has 0 prime factors, which sum to 0 (the empty sum). 0 is even, so 1 is in the sequence.
		

Links

• Eric Weisstein's World of Mathematics, Group.
• Wikipedia, Polynomial ring.

Crossrefs

Cf. A001222, A001414, A003961, A059897, A307150.

Intersection of any 2 of A003159, A028260, A036349.

Other lists that have conditions on the number of odd prime factors: A046337, A072978.

Subsequences: A001248, A006094\{6}, A046315, A056913.

Programs

Mathematica

{1}~Join~Select[Range@1000,(s=Flatten[Table@@@FactorInteger[#]];And@@EvenQ@{Length@s,Total@s})&] (* Giorgos Kalogeropoulos, Jun 24 2021 *)

PARI

iseven(x) = ((x%2) == 0);
isok(m) = my(f=factor(m)); iseven(sum(k=1, #f~, f[k,1]*f[k,2])) && iseven(sum(k=1, #f~, f[k,2])); \\ Michel Marcus, Jun 24 2021

PARI

is(n) = bigomega(n)%2 == 0 && valuation(n, 2)%2 == 0 \\ David A. Corneth, Jun 24 2021

Python

from sympy import factorint
def ok(n):
    f = factorint(n)
    return sum(f.values())%2 == 0 and sum(p*f[p] for p in f)%2 == 0
print(list(filter(ok, range(1, 216)))) # Michael S. Branicky, Jun 24 2021

Formula

{a(n) : n >= 1} = {m >= 1 : A001222(m) mod 2 = A001414(m) mod 2 = 0}.

{A036349(n) : n >= 1} = {a(n) : n >= 1} U {2 * a(n) : n >= 1}.

{A028260(n) : n >= 1} = {a(n) : n >= 1} U {A307150(a(n)) : n >= 1}.

For odd prime p, {A003159(n) : n >= 1} = {a(n) : n >= 1} U {A059897(a(n), p) : n >= 1}.