A324036 Modified reduced Collatz map fs acting on positive odd integers.
1, 5, 1, 11, 7, 17, 3, 23, 13, 29, 5, 35, 19, 41, 7, 47, 25, 53, 9, 59, 31, 65, 11, 71, 37, 77, 13, 83, 43, 89, 15, 95, 49, 101, 17, 107, 55, 113, 19, 119, 61, 125, 21, 131, 67, 137, 23, 143, 73, 149, 25, 155, 79, 161, 27, 167, 85, 173, 29, 179, 91, 185, 31, 191, 97, 197, 33, 203, 103, 209
Offset: 0
Examples
Iteration of fs on 11: 11, 17, 13, 3, 5, 1, whereas for fCr: 11, 17, 13 , 5, 1. The additional step (N = 1) occurs for 13 == 5 (mod 8), and 13 does not belong to the O(N;k) sets for N >= 2. The first additional N = 2 steps occur for 53 = a(26): 53, 13, 3, 5, 1, versus iteration of fCr: 53, 5, 1. Such N = 2 steps occur precisely after 53 + 64*k as 13 + 16*k and 3 + 4*k. The first additional N = 3 steps occur for 213 = a(106): 213, 53, 13, 3, 5, 1 versus 213, 5, 1 for fCr. The first additional N = 4 steps occur for 853 = a(426): 853, 213, 53, 13, 3, 5, 1 versus 853, 5, 1 for fCr.
Links
- Wolfdieter Lang, Collatz Trees from Vaillant-Delarue Maps
- Nicolas Vaillant and Philippe Delarue, The hidden face of the 3x+1 problem. Part I: Intrinsic algorithm, April 26 2019.
Programs
-
PARI
a(n) = my(m=Mod(n,4)); if (m==0, (2 + 3*n)/2, if (m==2, n/2, 2 + 3*n)); \\ Michel Marcus, Aug 10 2023
Formula
a(n) = fs(1 + 2*n) = (2 + 3*n)/2 if n == 0 (mod 4), a(n) = 2 + 3*n, for n == 1 or 3 (mod 4), and a(n) = n/2 if n == 2 (mod 4). This corresponds to fs(1 + 8*k) = 1 + 6*k, fs(3 + 8*k) = 5 + 12*k, fs(5 + 8*k) = 1 + 2*k, and fs(7 + 8*k) = 11 + 12*k, for k >= 0.
Conjectures from Colin Barker, Oct 14 2019: (Start)
G.f.: (1 + 5*x + x^2 + 11*x^3 + 5*x^4 + 7*x^5 + x^6 + x^7) / ((1 - x)^2*(1 + x)^2*(1 + x^2)^2).
a(n) = 2*a(n-4) - a(n-8) for n>7.
(End)
Extensions
More terms from Michel Marcus, Aug 10 2023
Comments