
COMMENTS

Corresponding first four values of 3*n + 1 are 5^4 * 11^4, 5^4 * 19^4, 5^4 * 43^4, 5^4 * 67^4.
Primes p such that both (5*p)^4 and ((5*p)^4  1)/3 are refactorable numbers begin 11, 19, 43, 67, 83, 109, 173, 211, 227, 443, 467, 557, 563, 587, 659, 739, 787, 821, 829, 853, 1123, 1187, 1229, 1277, 1453, 1523, 1571, 1709, 1901, 1973, 2083, 2099, 2237, 2467, 2531, 2621, 2909, 3347, 3517, 3877, 3923, 4099, 4243, 4253, 4259, 4483, 4547, ...; for each, p == 3 or 5 (mod 8).  Jon E. Schoenfield, Jan 21 2017
From Altug Alkan, Jan 25 2017: (Start)
Although numbers of the form ((5*p)^4  1)/3 appear in the beginning of sequence, note that not all terms are of the form ((5*p)^4  1)/3, i.e., (6239^161)/3.
However we can show that all terms are of the form 8 * A001318(m).
Proof: If an odd number n is in this sequence, then n must be a square and 3*n + 1 = 3 * (2*k + 1)^2 + 1 = 12 * k * (k + 1) + 4 = 24 * A000217(k) + 4 is a refactorable number. 4 = 2^2 is the highest power of 2 that divides 24 * A000217(k) + 4 because 6 * A000217(k) + 1 is an odd number. Since 24 * A000217(k) + 4 is not divisible by 3, 3*n + 1 cannot be a refactorable number when n is an odd refactorable number.
Since we proved that n is an even number, 3 * n + 1 is odd and it must be a square. If 3 * n + 1 = (2 * t + 1)^2, then n = ((2 * t + 1)^2  1) / 3 = 4 * t * (t + 1) / 3 = 8 * A001318(m). (End)
