A255138 - OEIS

A255138

a(n) = (1 + 2^n*(3 + 2*(-1)^n))/3.

2, 1, 7, 3, 27, 11, 107, 43, 427, 171, 1707, 683, 6827, 2731, 27307, 10923, 109227, 43691, 436907, 174763, 1747627, 699051, 6990507, 2796203, 27962027, 11184811, 111848107, 44739243, 447392427, 178956971

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OFFSET

0,1

COMMENTS

Let N_1 be the set of odd natural numbers and v(y) the 2-adic valuation of y. Define the map F : N_1 -> N_1 by F(x) = (3*x+1)/2^v(3*x+1) (see A075677). Let F^(k)(x) denote k-fold iteration of F, with recurrence F^(k)(x) = F(F^(k-1)(x)), k > 0, and initial condition F^(0)(x) = x. Then, for n>0, a(n) is the least m such that F^(n)(4*m-3) == 1 (mod 4). Cf. A257499.

Let k == 1 mod 4, and k(r) be the r-th iteration at which k appears in a Collatz sequence. When n >= 2 and k(r) == [2^(n+1) - a(n)] mod 2^(n+1), then n is the number of halving steps following k(r+1). For instance, since a(5) = 11, there are 5 halving steps following k(r+1) when k(r) == 53 mod 64, because 2^(5+1) = 64 and 64-11 = 53; e.g., k(r) = 117: 117 -> 352 -> 176 -> 88 -> 44 -> 22 -> 11. - Bob Selcoe, Feb 09 2017

LINKS

Table of n, a(n) for n=0..29.

Index entries for linear recurrences with constant coefficients, signature (1,4,-4).

FORMULA

a(2*n) = A136412(n); a(2*n+1) = A007583(n).

G.f.: (2-x-2*x^2)/((x-1)*(2*x-1)*(2*x+1)). - R. J. Mathar, Jul 25 2015

a(n) = a(n-1) + 4*a(n-2) - 4*a(n-3) for n > 2. - Wesley Ivan Hurt, Nov 05 2015

a(n) = 4*a(n-2) - 1. - Bob Selcoe, Feb 09 2017

a(n) = 2^(n+1) - A096773(n+1). - Ruud H.G. van Tol, Sep 04 2023

MAPLE

A255138:=n->(1 + 2^n*(3 + 2*(-1)^n))/3: seq(A255138(n), n=0..50); # Wesley Ivan Hurt, Nov 05 2015

MATHEMATICA