

A182137


Size of the set of b for numbers of the form 2^n*x + b that cannot be the smallest element of a set giving a duration of infinite flight in the Collatz problem.


1



1, 3, 6, 13, 28, 56, 115, 237, 474, 960, 1920, 3870, 7825, 15650, 31473, 63422, 126844, 254649, 509298, 1021248, 2050541, 4101082, 8219801, 16490635, 32981270, 66071490, 132455435, 264910870, 530485275, 1060970550, 2123841570, 4253619813, 8507239626, 17027951548, 34095896991, 68191793982, 136383587964, 272943149762, 546202840156, 1093108792776
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

In the Collatz Problem A014682, it is possible to apply the algorithm to first degree polynomials like 2^n*x+b, where n is integer and 0 <= b < 2^n. The iteration terminates by two cases:
1) a*x+b where a < 2^n: the polynomial is "minimized"
2) a*x+b where a is odd and a > 2^n, parity cannot be found. The polynomial cannot be minimized.
The sequence counts how many first degree polynomials end like first case for each n > 0.
The interest of this sequence is that every number that can be described by a minimized polynomial cannot be the smallest element of a set of value of T(n) = infinity.


LINKS

Table of n, a(n) for n=1..40.


EXAMPLE

Example with 4x+b (0 <= b < 4):
4x is even, thus gives 2x, 2 < 4 (first case).
4x+1, is odd thus 3(4x+1)+1 = 12x+4 is even, thus (12x+4)/2/2=3x+1 3 < 4, first case.
4x+2 is even, (4x+2)/2=2x+1, 2 < 4, first case.
4x+3 with same way gives 9x+5. 9 is odd and 9 > 4, second case.
That explains why the second (n=2) term in sequence is 3.


MATHEMATICA

a[n_] := Module[{b, p0, p1, minimized = 0}, For[b = 1, b <= 2^n, b++, {p0, p1} = {b, 2^n}; While[Mod[p1, 2] == 0 && p1 >= 2^n, {p0, p1} = If[Mod[p0, 2] == 0, {p0/2, p1/2}, {3*p0+1, 3*p1}]; If[p1<2^n, minimized += 1]]]; minimized]; Table[Print[an = a[n]]; an, {n, 1, 40}] (* JeanFrançois Alcover, Feb 12 2014, translated from D. S. McNeil's Sage code *)


PROG

(Sage)
def A182137(n):
....minimized = 0
....for b in xrange(2**n):
........p = [b, 2**n]
........while p[1] % 2 == 0 and p[1] >= 2**n:
............p[0], p[1] = [p[0]/2, p[1]/2] if p[0] % 2 == 0 else [3*p[0]+1, 3*p[1]]
........if p[1] < 2**n: minimized += 1
....return minimized # D. S. McNeil, Apr 14 2012


CROSSREFS

Cf. A074473, A186109.
Sequence in context: A032253 A125777 A103788 * A106461 A095768 A002478
Adjacent sequences: A182134 A182135 A182136 * A182138 A182139 A182140


KEYWORD

nonn


AUTHOR

Jérôme STORTI, Apr 14 2012


EXTENSIONS

More terms from D. S. McNeil, Apr 14 2012
One additional term from Jérôme STORTI, Apr 22 2012
a(39) from Jérôme STORTI, Jul 26 2012
a(40) from Jérôme STORTI, Feb 08 2014


STATUS

approved



