A101229 Perfect inverse "3x+1 conjecture" (See comments for rules).
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
Offset: 1
Examples
The first 4 is followed by 1 because 4 = 3*1 + 1, so rule 2: (4-1)/3 = 1; the second 4 is followed by 8 because the 2nd rule has already been applied, so rule 1: 4*2 = 8.
References
- R. K. Guy, Collatz's Sequence, Section E16 in Unsolved Problems in Number Theory, 2nd ed. New York: Springer-Verlag, pp. 215-218, 1994.
Links
- G. C. Greubel, Table of n, a(n) for n = 1..1000
- Eric Weisstein's World of Mathematics, Collatz Problem
- Index entries for linear recurrences with constant coefficients, signature (2).
Programs
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Magma
R
:=PowerSeriesRing(Integers(), 45); Coefficients(R!( x*(17*x^10+27*x^8+7*x^3-1)/(2*x-1) )); // G. C. Greubel, Mar 20 2019 -
Mathematica
Rest[CoefficientList[Series[x*(17*x^10+27*x^8+7*x^3-1)/(2*x-1), {x, 0, 45}], x]] (* G. C. Greubel, Mar 20 2019 *) LinearRecurrence[{2},{1,2,4,1,2,4,8,16,5,10,3},40] (* Harvey P. Dale, May 06 2023 *)
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PARI
my(x='x+O('x^45)); Vec(x*(17*x^10+27*x^8+7*x^3-1)/(2*x-1)) \\ G. C. Greubel, Mar 20 2019
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Sage
a=(x*(17*x^10+27*x^8+7*x^3-1)/(2*x-1)).series(x, 45).coefficients(x, sparse=False); a[1:] # G. C. Greubel, Mar 20 2019
Formula
a(n) = 3*2^(n-11) = 2^(n-11) + 2^(n-10) for n >= 11. - Jaroslav Krizek, Aug 17 2009
From Colin Barker, Apr 28 2013: (Start)
a(n) = 2*a(n-1) for n>11.
G.f.: x*(17*x^10+27*x^8+7*x^3-1) / (2*x-1). (End)
Extensions
More terms from Joshua Zucker, May 18 2006
Edited by G. C. Greubel, Mar 20 2019
Comments