

A075677


Reduced Collatz function R applied to the odd integers: a(n) = R(2n1), where R(k) = (3k+1)/2^r, with r as large as possible.


37



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
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OFFSET

1,2


COMMENTS

The evenindexed terms a(2i+2) = 6i+5 = A016969(i), i >= 0 [Comment corrected by Bob Selcoe, Apr 06 2015]. The oddindexed 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(4i1) = a(i). No multiple of 3 is in this sequence. See A075680 for the number of iterations of R required to yield 1.
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)
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

R. K. Guy, Unsolved Problems in Number Theory, E16.
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.
J. C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010; see p. 57, also (909), p. 306.


LINKS



FORMULA

For all n>=1 and for every k, there exists j>=0 dependent upon n and k such that either:
Eq. 1: a(n) = (3n1)/2^(2j+1) when k = ((4^(j+1)1)/3) mod 2^(2j+3). Alternatively: a(n) = A016789(n1)/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) = (3n1)/4^j when k = (5*2^(2j+1)  1)/3 mod 4^(j+1). Alternatively: a(n) = A016789(n1)/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.
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


EXAMPLE

a(11) = 1 because 21 is the 11th odd number and R(21) = 64/64 = 1.
i) 1 (mod 6) entry 1 = A016921(0) appears for n = A178415(1, k) = A347834(1, k1) (the arrays), for k >= 1, that is, for {1, 5, 21, ..} = A002450.
ii) 5 (mod 6) entry 11 = A007528(2) appears for n = A178415(4, k) = A347835(3, k1) (the arrays), for k >= 1, that is, for {7, 29, 117, ..} = A072261. (End)


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;


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

(Haskell)
(Python)
from sympy import divisors
def a(n):
return max(d for d in divisors(n) if d % 2)


CROSSREFS

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


KEYWORD

easy,nonn


AUTHOR



STATUS

approved



