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The terms are consecutive sextuples, ordered so that (A) a(6i-5) < a(6i-4) < ... < a(6i) for i > 0, and (B) a(6i+1) < a(6i+7) for i >= 0. This sequence has primitive solutions only. If k is relatively prime to all of the terms in a primitive sextuple, then multiplying the terms in that sextuple by k gives another solution - see A322682.

The terms are consecutive triples, ordered so that (A) a(3i-2) < a(3i-1) < a(3i), and (B) a(3i-2) < a(3j-2) for i < j. This sequence has primitive terms only. If k is relatively prime to all of the terms in a primitive triple, then multiplying the terms in that triple by k gives another solution - see A322679.

The terms are consecutive quadruples, ordered so that (A) a(4i-3) < a(4i-2) < a(4i-1) < a(4i) for i > 0, and (B) a(4i+1) < a(4i+5) for i >= 0. This sequence has primitive terms only. If k is relatively prime to all of the terms in a primitive quadruple, then multiplying the terms in that quadruple by k gives another solution - see A322680.

The terms are consecutive quintuples, ordered so that (A) a(5i-4) < a(5i-3) < ... < a(5i) for i > 0, and (B) a(5i+1) < a(5i+6) for i >= 0. This sequence has primitive terms only. If k is relatively prime to all of the terms in a primitive quintuple, then multiplying the terms in that quintuple by k gives another solution - see A322681.

From David A. Corneth, Feb 15 2019: (Start)

Some numbers occur in more than one quintuple, for example 1773744050 is in the quintuples [1579877800, 1652932372, 1653851276, 1663815260, 1773744050] and [1652932372, 1653851276, 1663815260, 1773744050, 1774581050].

The 4693 distinct terms in the first 5000 terms have only 111 distinct prime factors, the largest being 22751. All of these primes differ 1 from a 29-smooth number. (End)

From David A. Corneth, Feb 17 2019: (Start)

A quintuple (e1, e2, e3, e4, e5) is valid and primitive if and only if

1. The elements are in increasing order.

2. Every element e of the quintuple has the same value for phi(e), sigma(e) and tau(e).

3. For every number k between e1 and e5 that's not in the quintuple, at least one of the following statements is false: phi(e1) = phi(k), sigma(e1) = sigma(k), tau(e1) = tau(k).

4. Let g be gcd(e1, e2, e3, e4, e5). Then for every d|g, (e1/d, e2/d, e3/d, e4/d, e5/d) is not a valid quintuple. Therefore, (e1, e2, e3, e4, e5) is primitive. (End)

The terms are consecutive septuples, ordered so that (A) a(7i-6) < a(7i-5) < ... < a(7i) for i > 0, and (B) a(7i+1) < a(7i+8) for i >= 0. This sequence has primitive solutions only. If k is relatively prime to all of the terms in a primitive septuple, then multiplying the terms in that septuple by k gives another solution - see A322683.

The terms are consecutive octuples, ordered so that (A) a(8i-7) < a(8i-6) < ... < a(8i) for i > 0, and (B) a(8i+1) < a(8i+9) for i >= 0. This sequence has primitive solutions only. If k is relatively prime to all of the terms in a primitive octuple, then multiplying the terms in that octuple by k gives another solution - see A322684.

The terms are consecutive 10-tuples, ordered so that (A) a(10i-9) < a(10i-8) < ... < a(10i) for i > 0, and (B) a(10i+1) < a(10i+11) for i >= 0. This sequence has primitive solutions only. If k is relatively prime to all of the terms in a primitive 10-tuple, then multiplying the terms in that 10-tuple by k gives another solution - see A322686.

The terms are consecutive 11-tuples, ordered so that (A) a(11i-10) < a(11i-9) < ... < a(11i) for i > 0, and (B) a(11i+1) < a(11i+12) for i >= 0. Primitive solutions are in A322697.

The terms are consecutive pairs, ordered so that (A) a(2i-1) < a(2i) for i > 0, and (B) a(2i+1) < a(2i+3) for i >= 0. This sequence has primitive solutions only. If k is relatively prime to all of the terms in a primitive pair, then multiplying the terms in that pair by k gives another solution - see A134922. In Burton's book (see references), problem 3 in section 7.2 asks the reader to prove a special case for (568,638).

The terms are consecutive 9-tuples, ordered so that (A) a(9i-8) < a(9i-7) < ... < a(9i) for i > 0, and (B) a(9i+1) < a(9i+10) for i >= 0. This sequence has primitive solutions only. If k is relatively prime to all the terms in a primitive 9-tuple, then multiplying the terms in that 9-tuple by k gives another solution - see A322685.