OFFSET
1,2
COMMENTS
Construction by subgroup generation: (Start)
The set of numbers congruent to 1 modulo 8 (A017077) contains all the odd squares and generates a subgroup of the positive rational numbers (under multiplication) that contains no additional integers. The subgroup has an infinite number of cosets. The rest of the construction process extends the subgroup, reducing the number of cosets to 2, by choosing additional generators that are semiprime.
First we extend the subgroup to include all nonzero integer squares. As we already have the odd squares, we need only add 4, the square of the smallest prime, as a generator. The extended subgroup has only 8 cosets and its integer members are listed in A234000. To achieve a subgroup with 2 cosets we now add squarefree semiprime generators. The 2 smallest, 6 and 10, suffice.
The resulting subgroup has this sequence's terms as its integer members.
(End)
The equivalent process starting with numbers congruent to 1 modulo 3 (or 1 modulo 6) produces A189715. If we take its intersection with this sequence we get A370268, which starts with the first 72 nonzero numbers of the form x^2 + 6y^2 (see A002481). Similarly, if we start with numbers congruent to 1 modulo 5 (or 1 modulo 10) and take the resulting set's intersection with this sequence we get a set starting with the first 32 nonzero numbers of the form x^2 - 10y^2 (see A242664).
The construction process leads to a number of properties:
- The sequence is closed under multiplication and all integer ratios between terms are in the sequence.
- The sequence and its complement have the property that the terms of one can be generated by halving the even terms of the other. Each has asymptotic density 1/2.
Numbers whose squarefree part is congruent to {1,7} mod 8 or {6,10} mod 16.
EXAMPLE
7 is prime, so 7 is its only prime factor, which has the form 8m-1. So 7 has an even number (zero) of prime factors not of the form 8m+-1, and therefore is in the sequence. In terms of the subgroup generators described at the start of the comments, (13*8+1) * 4 / (6*10) = 105 * 4/60 = 7.
110 = 2 * 5 * 11, so it has 3 prime factors and all 3 do not have the form 8m+-1. 3 is odd, so 110 is not in the sequence.
PROG
(PARI) isok(k) = {c = core(k); c%8 == 1 || c%8 == 7 || c%16 == 6 || c%16 == 10}
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Munn, Feb 13 2024
STATUS
approved