cp's OEIS Frontend

This sequence is a permutation of all numbers not congruent to 4 (mod 6) (A047256).

This sequence has no finite cycles other than (2,1) under recursion, because of all cycles of permutation A093545(n), only one cycle (2,1) is without any number congruent to 4 (mod 6). See section "formula" first entry here. After a finite number of recursions it will reach only numbers divisible by 3 under further recursion.

m = a(n) is the smallest solution to A014682(m) = n.

If we define f(n) = 2*n and j(n) as an arbitrary recursion into a(n) and/or f(n) ( two examples: j(n) = a(a(n)) or j(n) = a(f(a(n))) ), then for all m, k and n = A014682^k(m), exists a j(n) that allows m = j(n). "^k" means recursion here.

Proving the Collatz conjecture could be done by proving that for all positive integers m, a function j(n) (see first comment) could be designed such that m = j(1). All numbers greater than 4 can be reached by a^k(6*p - 2) with exactly one p and k. The Collatz conjecture cannot be true if 3*p - 1 = a^k(6*p - 2) exists.

This sequence occurs twice as a linear spoke in the hexagonal spiral constructed from A002266:

17 17 17 17 17 18 18

16 11 11 11 11 12 12 18

16 11 6 6 7 7 7 12 18

16 10 6 3 3 3 3 7 12 18

16 10 6 3 1 1 1 4 7 12 19

16 10 6 2 0 0 0 1 4 8 13 19

15 10 5 2 0 0 1 4 8 13 19

15 10 5 2 2 2 4 8 13 19

15 9 5 5 5 4 8 13 19

15 9 9 9 9 8 13 20

15 14 14 14 14 14 20

a(-1-n) = 0, 1, 4, 8, 13, 19, 26, 35, 45, ... also occurs twice in the same spiral.

Difference table:

0, 1, 3, 6, 11, 17, 24, 32, 41, 52, ... = a(n)

1, 2, 3, 5, 6, 7, 8, 9, 11, 12, ... = A047256(n+1)

1, 1, 2, 1, 1, 1, 1, 2, 1, 1, ... = A130782.

There is no linear spoke with three copies in this spiral. Compare with the spiral illustrated in sequence A330707 and constructed from A002265 where the same spokes occur three times: A006578, A001859 and A077043, essentially. Strictly, three times from 1, 1, 1 for A006578, from 2, 2, 2 for A001859 and from 7, 7, 7 for A077043.