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A275544 Number of distinct terms at a given iteration of the Collatz (or 3x+1) map starting with 0. 1
1, 2, 4, 8, 15, 29, 56, 108, 208, 400, 766, 1465, 2793, 5314, 10088, 19115, 36156, 68290, 128817, 242720, 456884, 859269, 1614809, 3032673, 5692145, 10678326, 20023239, 37531218, 70323203, 131725663, 246674211 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

If one considers an algebraic approach to the Collatz conjecture, the tree of outcomes of the "Half Or Triple Plus One" process starting with a natural number n:

  i

  0:                                   n

  1:                  3n+1                            n/2

  2:        9n+4        (3/2)n+1/2         (3/2)n+1          n/4

  3: 27n+13 (9/2)n+2 (9/2)n+5/2 (3/4)n+1/4 (9/2)n+4 (3/4)n+1/2 (3/4)n+1 n/8

  ...

  reveals that any n that is part of a cycle has to satisfy an equation of the following form:

  (3^(i-p)/2^p - 1)n + x_i = 0     i = 0,1,2,3,...  p = 0,...,i

  where x_i are the possible constant terms at iteration i, ie.,

  x_0 = [0],

  x_1 = [1,0],

  x_2 = [4,1/2,1,0],

  x_3 = [13,2,5/2,1/4,4,1/2,1,0],

  x_4 = [40,13/2,7,1,17/2,5/4,7/4,1/8,13,2,5/2,1/4,4,1/2,1,0],

  ...

(Note that not all the combinations of members of x_i and numbers p yield an equation that corresponds to n having to belong to a cycle, instead satisfying at least one equation of the form above is a necessary condition for every n that does).

This sequence is composed of the numbers of distinct possible constant terms at each iteration i.

The only constant term at the zeroth iteration is 0. Since at each iteration both half and triple plus one is considered, the halving of 0 always yields another 0, which always has the same progression tree, and therefore each set x_i contains the members of all previous sets x_j where j < i. This is also the reason why the sequence at the beginning resembles powers of 2 A000079, but later falls behind as more and more duplicates arise.

This sequence is related to A275545, if one sequence is known it is possible to work out the other (see formula).

An empirical observation suggests that the same sequence of numbers arises if we analogously consider the 3n-1 problem (the Collatz conjecture can be referred to as the 3n+1 problem).

The first 9 terms coincide with the Tetranacci numbers A000078.

LINKS

Table of n, a(n) for n=0..30.

Wikipedia, Collatz conjecture

FORMULA

a(0) = 1; a(n) = 2*a(n-1) - A275545(n), n >= 1.

EXAMPLE

a(3) = 8 since x_3 has 8 members and they are all distinct.

a(4) = 15 since x_4 has 16 members but the number 1 appears twice.

PROG

(Python)

x = [0]

for i in range(n):

    x_tmp = []

    for s in x:

        x_tmp.append(3*s+1)

        x_tmp.append(s*0.5)

    x = x_tmp

    x = list(set(x))

    print len(x)

(Python)

from fractions import Fraction

A275544_list, c = [1], [Fraction(0, 1)]

for _ in range(20):

    c = set(e for d in c for e in (3*d+1, d/2))

    A275544_list.append(len(c)) # Chai Wah Wu, Sep 02 2016

CROSSREFS

Cf. A275545.

Sequence in context: A108564 A066369 A239555 * A000078 A176503 A262333

Adjacent sequences:  A275541 A275542 A275543 * A275545 A275546 A275547

KEYWORD

nonn,more

AUTHOR

Rok Cestnik, Aug 01 2016

EXTENSIONS

a(27)-a(29) corrected and a(30) added by Chai Wah Wu, Sep 02 2016

STATUS

approved

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Last modified February 20 14:53 EST 2018. Contains 299380 sequences. (Running on oeis4.)