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Reduced Collatz function R applied to the odd integers: a(n) = R(2n-1), where R(k) = (3k+1)/2^r, with r as large as possible.
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%I #127 Aug 26 2024 04:58:03

%S 1,5,1,11,7,17,5,23,13,29,1,35,19,41,11,47,25,53,7,59,31,65,17,71,37,

%T 77,5,83,43,89,23,95,49,101,13,107,55,113,29,119,61,125,1,131,67,137,

%U 35,143,73,149,19,155,79,161,41,167,85,173,11,179,91,185,47,191,97,197

%N Reduced Collatz function R applied to the odd integers: a(n) = R(2n-1), where R(k) = (3k+1)/2^r, with r as large as possible.

%C The even-indexed terms a(2i+2) = 6i+5 = A016969(i), i >= 0 [Comment corrected by _Bob Selcoe_, Apr 06 2015]. The odd-indexed terms terms are the same as A067745. Note that this sequence is A016789 with all factors of 2 removed from each term. Also note that a(4i-1) = a(i). No multiple of 3 is in this sequence. See A075680 for the number of iterations of R required to yield 1.

%C From _Bob Selcoe_, Apr 06 2015: (Start)

%C All numbers in this sequence appear infinitely often.

%C From Eq. 1 and Eq. 2 in Formulas: Eq. 1 is used with 1/3 of the numbers in this sequence, Eq. 2 is used with 2/3 of the numbers.

%C (End)

%C Empirical: For arbitrary m, Sum_{n=2..A007583(m)} (a(n) - a(n-1)) = 0. - _Fred Daniel Kline_, Nov 23 2015

%C From _Wolfdieter Lang_, Dec 07 2021: (Start)

%C Only positive numbers congruent to 1 or 5 modulo 6 appear.

%C i) For the sequence entry with value A016921(m), for m >= 0, that is, a value from {1, 7, 13, ...}, the indices n are given by the row of array A178415(2*m+1, k), for k >= 1.

%C ii) For the sequence entry with value A007528(m), for m >= 1, that is, a value from {5, 11, 17, ...}, the indices n are given by the row of array A178415(2*m, k), for k >= 1.

%C See also the array A347834 with permuted row numbers and columns k >= 0. (End)

%D Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section E16, pp. 330-336.

%D Victor Klee and Stan Wagon, Old and new unsolved problems in plane geometry and number theory, The Mathematical Association of America, 1991, p. 225, C(2n+1) = a(n+1), n >= 0.

%D Jeffrey C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010; see p. 57, also (90-9), p. 306.

%H T. D. Noe, <a href="/A075677/b075677.txt">Table of n, a(n) for n = 1..1000</a>

%H Marc Chamberland, <a href="http://www.math.grinnell.edu/~chamberl/papers/3x_survey_cat.pdf">Una actualizacio del problema 3x + 1</a>, Butl. Soc. Catalana Mat. (18), 19-45, 2003.

%H Jeffrey C. Lagarias, <a href="http://www.cecm.sfu.ca/organics/papers/lagarias/paper/html/paper.html">The 3x+1 problem and its generalizations</a>, Amer. Math. Monthly, 92 (1985), 3-23.

%H Fabian S. Reid, <a href="https://arxiv.org/abs/2105.07955">The Visual Pattern in the Collatz Conjecture and Proof of No Non-Trivial Cycles</a>, arXiv:2105.07955 [math.GM], 2021.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CollatzProblem.html">Collatz Problem</a>.

%H <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>.

%F a(n) = A000265(6*n-2) = A000265(3*n-1). - _Reinhard Zumkeller_, Jan 08 2014

%F From _Bob Selcoe_, Apr 05 2015: (Start)

%F For all n>=1 and for every k, there exists j>=0 dependent upon n and k such that either:

%F Eq. 1: a(n) = (3n-1)/2^(2j+1) when k = ((4^(j+1)-1)/3) mod 2^(2j+3). Alternatively: a(n) = A016789(n-1)/A081294(j+1) when k = A002450(j+1) mod A081294(j+2). Example: n=51; k=101 == 5 mod 32, j=1. a(51) = 152/8 = 19.

%F or

%F Eq. 2: a(n) = (3n-1)/4^j when k = (5*2^(2j+1) - 1)/3 mod 4^(j+1). Alternatively: a(n) = A016789(n-1)/A000302(j) when k = A072197(j) mod A000302(j+1). Example: n=91; k=181 == 53 mod 64, j=2. a(91) = 272/16 = 17.

%F (End) [Definition corrected by _William S. Hilton_, Jul 29 2017]

%F a(n) = a(n + g*2^r) - 6*g, n > -g*2^r. Examples: n=59; a(59)=11, r=5. g=-1: 11 = a(27) = 5 - (-1)*6; g=1: 11 = a(91) = 17 - 1*6; g=2: 11 = a(123) = 23 - 2*6; g=3: 11 = a(155) = 29 - 3*6; etc. - _Bob Selcoe_, Apr 06 2015

%F a(n) = a((1 + (3*n - 1)*4^(k-1))/3), k>=1 (cf. A191669). - _L. Edson Jeffery_, Oct 05 2015

%F a(n) = a(4n-1). - _Bob Selcoe_, Aug 03 2017

%F a(n) = A139391(2n-1). - _Antti Karttunen_, May 06 2024

%F Sum_{k=1..n} a(k) ~ n^2. - _Amiram Eldar_, Aug 26 2024

%e a(11) = 1 because 21 is the 11th odd number and R(21) = 64/64 = 1.

%e From _Wolfdieter Lang_, Dec 07 2021: (Start)

%e i) 1 (mod 6) entry 1 = A016921(0) appears for n = A178415(1, k) = A347834(1, k-1) (the arrays), for k >= 1, that is, for {1, 5, 21, ..} = A002450.

%e ii) 5 (mod 6) entry 11 = A007528(2) appears for n = A178415(4, k) = A347835(3, k-1) (the arrays), for k >= 1, that is, for {7, 29, 117, ..} = A072261. (End)

%p f:=proc(n) local t1;

%p if n=1 then RETURN(1) else

%p t1:=3*n+1;

%p while t1 mod 2 = 0 do t1:=t1/2; od;

%p RETURN(t1); fi;

%p end;

%p # _N. J. A. Sloane_, Jan 21 2011

%t nextOddK[n_] := Module[{m=3n+1}, While[EvenQ[m], m=m/2]; m]; (* assumes odd n *) Table[nextOddK[n], {n, 1, 200, 2}]

%t v[x_] := IntegerExponent[x, 2]; f[x_] := (3*x + 1)/2^v[3*x + 1]; Table[f[2*n - 1], {n, 66}] (* _L. Edson Jeffery_, May 06 2015 *)

%o (PARI) a(n)=n+=2*n-1;n>>valuation(n,2) \\ _Charles R Greathouse IV_, Jul 05 2013

%o (Haskell)

%o a075677 = a000265 . subtract 2 . (* 6) -- _Reinhard Zumkeller_, Jan 08 2014

%o (Python)

%o from sympy import divisors

%o def a(n):

%o return max(d for d in divisors(n) if d % 2)

%o print([a(6*n - 2) for n in range(1, 101)]) # _Indranil Ghosh_, Apr 15 2017, after formula by _Reinhard Zumkeller_

%Y Cf. A000265, A000302, A002450, A007528, A016789, A016921, A016969, A065677, A072197, A072261, A075680, A081294, A178415, A191669, A329480, A347834.

%Y Cf. A006370, A014682 (for non-reduced Collatz maps), A087230 (A371093), A371094.

%Y Odd bisection of A139391.

%Y Even bisection of A067745, which is also the odd bisection of this sequence.

%Y After the initial 1, the second leftmost column of A256598.

%Y Row 2 of A372283.

%K easy,nonn

%O 1,2

%A _T. D. Noe_, Sep 25 2002