cp's OEIS Frontend

For each prime P check the generalized Collatz sequence of each integer N > 1 defined by c(1) = N, c(n+1) = c(n) * P + 1 if F > P, otherwise c(n+1) = c(n) / F, where F is the smallest factor of c(n), until c(n) = c(m) for n > m starts a cycle. If all c(i) > 1, then P does not belong to the sequence (and vice versa).

All terms are as yet only conjectures. Jeff Heleen checked the primes < 1000 and start points up to 10000000 (see Prime Puzzle 114 and example below). a(1)=3 is the ordinary Collatz problem. - Frank Ellermann, Jan 20 2002

The jOEIS program uses start points up to 10^8 and yields [3, 5, 7, 19, 29, 41, 43*, 53, 71*, 79, 89*, 103*, 107, 109*, 127, 131*, 137] followed by [139, 149, 157, 179, 191, 197, 199, 211, 227, ...]. The terms in the first list without asterisks agree with A106919. - Georg Fischer, Jun 17 2023

Let f(x) be any monic integral cubic polynomial with discriminant -n and irreducible over Z. Consider the set S of primes p such that f(x) has no zeros in Zp, i.e., f(x) is irreducible in Zp. For the discriminants -n in this sequence, set S coincides with the primes represented by one binary quadratic form ax^2+bxy+cy^2 with -n=b^2-4ac. For examples, see A106867, A106872, A106282, A106919, A106954, A106967, A040034 and A040038. This sequence consists of (1) terms 4d in A106312 such that the class number of d is 1, (2) terms d in A106312 such that the class number of d is 3 and (3) 108 and 243.