

A101229


Perfect inverse "3x+1 conjecture": rule 1: multiply n by 2 to give n' = 2n. rule 2: when n'=(3x+1), do n"= (n'1)/3 (n" integer) Additional rule: rule 2 is applied once for any number n' (otherwise, the sequence beginning with 1 would be the cycle "1 2 4 1 2 4 1 2 4 1..."); then apply rule 1.


2



1, 2, 4, 1, 2, 4, 8, 16, 5, 10, 3, 6, 12, 24, 48, 96, 192, 384, 768, 1536, 3072, 6144, 12288, 24576, 49152, 98304, 196608, 393216, 786432, 1572864, 3145728, 6291456, 12582912, 25165824, 50331648, 100663296, 201326592, 402653184, 805306368
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OFFSET

1,2


COMMENTS

Gives a particular sequence of hailstone numbers which, perhaps, may be considered as a central axis for all the hailstone numbers sequences. The perfect inverse "3x+1 conjecture" falls rapidly into the sequence 3 6 12 24 48 96... which will never give a number to which apply the 2nd rule.
a(n) for n >= 11 written in base 2: 11, 110, 11000, 110000, ..., i.e.: 2 times 1, (n11) times 0 (see A003953(n10). [From Jaroslav Krizek, Aug 17 2009]


REFERENCES

R. K. Guy, Collatz's Sequence, Section E16 in Unsolved Problems in Number Theory, 2nd ed. New York: SpringerVerlag, pp. 215218, 1994.


LINKS

Table of n, a(n) for n=1..39.
Eric Weisstein's World of Mathematics, Collatz Problem
Index entries for linear recurrences with constant coefficients, signature (2).


FORMULA

a(n) = 3*2^(n11) = 2^(n11) + 2^(n10) for n >= 11. [Jaroslav Krizek, Aug 17 2009]
a(n) = 2*a(n1) for n>11. G.f.: x*(17*x^10+27*x^8+7*x^31) / (2*x1).  Colin Barker, Apr 28 2013


EXAMPLE

The first 4 is followed by 1 because 4=3*1+1, so rule 2: (41)/3=1;
the second 4 is followed by 8 because the 2nd rule has already been applied, so rule 1: 4x2=8


CROSSREFS

Cf. A070165, A006577, A006667, A006666, A070167.
Sequence in context: A115314 A062039 A035492 * A057176 A194671 A131398
Adjacent sequences: A101226 A101227 A101228 * A101230 A101231 A101232


KEYWORD

nonn,uned,easy


AUTHOR

Alexandre Wajnberg, Jan 22 2005


EXTENSIONS

More terms from Joshua Zucker, May 18 2006


STATUS

approved



