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A075677 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. 27

%I

%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)} (A075677(n) - A075677(n-1)) = 0. - _Fred Daniel Kline_, Nov 23 2015

%D R. K. Guy, Unsolved Problems in Number Theory, E16.

%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 J. 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 M. 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 J. 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 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

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

%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): return max(list(filter(lambda i: i%2 == 1, divisors(n))))

%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, A016789, A016969, A065677, A072197, A075680, A081294.

%K easy,nonn

%O 1,2

%A _T. D. Noe_, Sep 25 2002

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Last modified January 21 22:55 EST 2020. Contains 331129 sequences. (Running on oeis4.)