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 A324036 Modified reduced Collatz map fs acting on positive odd integers. 4
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS This is a modification of the reduced Collatz map given in A075677. The Collatz conjecture is that iteration of the map fs leads to 1 for all positive odd integers. In the Vaillant-Delarue (V-D) reference the present map fs: Odd -> Odd, 2*n+1 -> a(n) = fs(2*n+1), for n >= 0, is called f_{s}. The differences from b(n) = A075677(n+1) = fCr(2*n+1) (called f_{cr} in V-D) occur for the positions n = 2 + 4*k, for k >= 1: b(2 + 4*k) = b(k) = A075677(k+1) but a(2 + 4*k) = 1 + 2*k, which differs. The advantage of the map fs (or a) over fCr (or b) is an explicit formula over a recurrence. Additional steps are introduced in the iteration of fs versus fCr. This leads to an incomplete binary tree, called CfsTree, given in A324038. No such tree is available for fCr. Such additional steps in fs can only occur after odd numbers congruent to 5 modulo 8: fs(5 + 8*k) = a(2 + 4*k) = 1 + 2*k  and fs(1 + 2*k) = a(k). On the other hand, fCr(5 + 8*k) = b(2 + 4*k) = b(k). The appearance of exactly N consecutive steps in fs versus fCr, for N >= 2, can be shown recursively to start with the odd numbers O(N;k) = 1 + 4*O(N-1;k), for N >= 3, with input O(2;k) = 53 + (4^3)*k. These are the numbers  O(N;k) = A072197(N) + A000302(N+1)*k, for N >= 2. Therefore only one additional step follows directly after an odd number 5 (mod 8) if it is not of the O(N;k) type for N >= 2. The minimal number of iterations of function fs acting on 2*n + 1  (or a acting on n), for n >= 0, to reach 1 is given in A324037 (if for very large n the number 1 should not be reached A324037(n) is set to -1). 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. 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) EXAMPLE 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. CROSSREFS Cf. A000302, A072197, A075677, A324037, A324038. Sequence in context: A242060 A185953 A323359 * A075677 A051853 A159074 Adjacent sequences:  A324033 A324034 A324035 * A324037 A324038 A324039 KEYWORD nonn,easy AUTHOR Nicolas Vaillant, Philippe Delarue, Wolfdieter Lang, May 08 2019 STATUS approved

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