OFFSET
1,1
COMMENTS
For each integer coefficient C >= 3, check the generalized Collatz sequence of each integer N > 1 defined by c(1) = N, c(n+1) = c(n) * C + 1 if F > C, otherwise c(n+1) = c(n) / F, where F is the smallest factor of c(n). If all integers reduce to 1, C belongs in the sequence.
A058047 is the primes in this sequence, as proposed by Zhongfu and Shiming; this sequence extends to consider composite C.
All terms are as yet only conjectures. Link includes github of code implementing CPU and GPU tests of these coefficients. Tests include reductions of integers up to 10^7 and random integers of size 2^2048.
LINKS
J. C. Lagarias The 3x + 1 Problem: An Annotated Bibliography, II (2000-2009).
John Laky, Python Collatz tests.
EXAMPLE
The first term in the sequence, 3, is the normal Collatz formula C_3 is defined as:
n / 2 if n is congruent to 0 mod 2,
otherwise, 3*n+1.
Note that 3 is the first element of this sequence iff the Collatz Conjecture is true.
The next term, 5, creates C_5 is defined as:
n / 2 if n is congruent to 0 mod 2,
n / 3 if n is congruent to 0 mod 3,
otherwise, 5*n+1.
However, the coefficient 11 is not in this sequence, as it creates the formula C_11:
n / 2 if n is congruent to 0 mod 2,
n / 3 if n is congruent to 0 mod 3,
n / 5 if n is congruent to 0 mod 5,
n / 7 if n is congruent to 0 mod 7,
otherwise, 11*n+1.
C_11 creates a nontrivial loop 188, 518, 408, 188. The loop is nontrivial because it does not contain 1. Thus, 11 does not belong in the sequence.
PROG
(Python)
import sympy
BIT_LIMIT = 100000
MAX_TESTS = 10000000
def _reduce(c, n):
seen = []
while n != 1:
for p in sympy.sieve.primerange(c):
while n % p == 0: n = n // p
if n==1: break
n = c*n+1
if n.bit_length() > BIT_LIMIT or n in seen: break
seen.append(n)
return n
def is_A374670(c):
for i in range(1, MAX_TESTS):
if _reduce(c, i) != 1: return False
return True
CROSSREFS
KEYWORD
nonn,more
AUTHOR
John Laky, Jul 14 2024
STATUS
approved