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)

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

From Wolfdieter Lang, Dec 07 2021: (Start)

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

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.

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.

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

REFERENCES

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

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.

Jeffrey 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

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

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

Fabian S. Reid, The Visual Pattern in the Collatz Conjecture and Proof of No Non-Trivial Cycles, arXiv:2105.07955 [math.GM], 2021.

Eric Weisstein's World of Mathematics, Collatz Problem.

FORMULA

From Bob Selcoe, Apr 05 2015: (Start)

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

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.

or

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.

(End) [Definition corrected by William S. Hilton, Jul 29 2017]

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

a(n) = a((1 + (3*n - 1)*4^(k-1))/3), k>=1 (cf. A191669). - L. Edson Jeffery, Oct 05 2015

a(n) = a(4n-1). - Bob Selcoe, Aug 03 2017

a(n) = A139391(2n-1). - Antti Karttunen, May 06 2024

Sum_{k=1..n} a(k) ~ n^2. - Amiram Eldar, Aug 26 2024

EXAMPLE

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

From Wolfdieter Lang, Dec 07 2021: (Start)

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;

# 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

(Python)

from sympy import divisors

def a(n):

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

print([a(6*n - 2) for n in range(1, 101)]) # Indranil Ghosh, Apr 15 2017, after formula by Reinhard Zumkeller

CROSSREFS

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

Odd bisection of A139391.

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

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

Row 2 of A372283.

KEYWORD

easy,nonn

AUTHOR

T. D. Noe, Sep 25 2002

STATUS

approved