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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. 10
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, 77, 5, 83, 43, 89, 23, 95, 49, 101, 13, 107, 55, 113, 29, 119, 61, 125, 1, 131, 67, 137, 35, 143, 73, 149, 19, 155, 79, 161, 41, 167, 85, 173, 11, 179, 91, 185, 47, 191, 97, 197 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

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.

From Bob Selcoe, Apr 06 2015 (Start):

All numbers in this sequence appear infinitely often.

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.

(End)

REFERENCES

M. Chamberland, Una actualizacio del problema 3x + 1, Butl. Soc. Catalana Mat. (18), 19-45, 2003.

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

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

LINKS

T. D. Noe, Table of n, a(n) for n = 1..1000

J. C. Lagarias, The 3x+1 problem and its generalizations, Amer. Math. Monthly, 92 (1985), 3-23.

Eric Weisstein's World of Mathematics, Collatz Problem

Index entries for sequences related to 3x+1 (or Collatz) problem

FORMULA

a(n) = A000265(6*n-2). - Reinhard Zumkeller, Jan 08 2014

From Bob Selcoe, Apr 05 2015 (Start):

For j>=0, the following two equations make a single function for all k:

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).

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).

(End)

a(n) = a(n + g*2^r) - 6*g. - Bob Selcoe, Apr 06 2015

EXAMPLE

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

From  Bob Selcoe, Apr 05 2015 (Start):

From Eq. 1: n=51; k=101 == 5 mod 32, j=1. a(51) = 152/8 = 19.

From Eq. 2: n=91; k=181 == 53 mod 64, j=2.  a(91) = 272/16 = 17.

(End)

n=59; a(59)=11, r=5. g = -1: 11 = a(27)=5 + 6; g = 1: 11 = a(91)=17 - 6; g=2: 11 = a(123)=23 - 2*6; g=3: 11 = a(155)=29 - 3*6; etc. - Bob Selcoe, Apr 06 2015

MAPLE

f:=proc(n) local t1;

if n=1 then RETURN(1) else

t1:=3*n+1;

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

RETURN(t1); fi;

end;

(from N. J. A. Sloane, Jan 21 2011)

MATHEMATICA

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

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 *)

PROG

(PARI) a(n)=n+=2*n-1; n>>valuation(n, 2) \\ Charles R Greathouse IV, Jul 05 2013

(Haskell)

a075677 = a000265 . subtract 2 . (* 6) -- Reinhard Zumkeller, Jan 08 2014

CROSSREFS

Cf. A000302, A002450, A016789, A016969, A065677, A072197, A075680, A081294.

Sequence in context: A131782 A242060 A185953 * A051853 A159074 A147414

Adjacent sequences:  A075674 A075675 A075676 * A075678 A075679 A075680

KEYWORD

easy,nonn

AUTHOR

T. D. Noe, Sep 25 2002

STATUS

approved

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Last modified May 25 13:33 EDT 2015. Contains 257808 sequences.