

A165355


a(n) = 3n + 1 if n is even, or a(n) = (3n + 1)/2 if n is odd.


12



1, 2, 7, 5, 13, 8, 19, 11, 25, 14, 31, 17, 37, 20, 43, 23, 49, 26, 55, 29, 61, 32, 67, 35, 73, 38, 79, 41, 85, 44, 91, 47, 97, 50, 103, 53, 109, 56, 115, 59, 121, 62, 127, 65, 133, 68, 139, 71, 145, 74, 151, 77, 157, 80, 163, 83, 169, 86, 175, 89, 181, 92, 187, 95, 193, 98
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OFFSET

0,2


COMMENTS

Second trisection of A026741.
A111329(n+1) = A000041(a(n)).  Reinhard Zumkeller, Nov 19 2009
We observe that this sequence is a particular case of sequence for which there exists q: a(n+3) = (a(n+2)*a(n+1)+q)/a(n) for every n >= n0. Here q=9 and n0=0.  Richard Choulet, Mar 01 2010
The entries are also encountered via the bilinear transform approximation to the natural log (unit circle). Specifically, evaluating 2(z1)/(z+1) at z = 2, 3, 4, ..., A165355 entries stem from the pair (sums) seen 2 ahead of each new successive prime. For clarity, the evaluation output is 2, 3, 1, 1, 6, 5, 4, 3, 10, 7, 3, 2, 14, 9, 8, 5, 18, 11, ..., where (1+1), (4+3), (3+2), (8+5), ... generate the A165355 entries (after the first). As an aside, the same mechanism links A165355 to A140777.  Bill McEachen, Jan 08 2015
As a followup to the previous comment, it appears that the numerators and denominators of 2(z1)/(z+1) are respectively given by A145979 and A060819, but with different offsets.  Michel Marcus, Jan 14 2015
When elements of this series are reduced by the highest possible power of two this series is equivalent to A067745. E.g.: 1, 2/2, 7, 5, 13, 8/8 ....  Joe Slater, Nov 30 2016


LINKS

G. C. Greubel, Table of n, a(n) for n = 0..1000
D. H. Lehmer, Continued fractions containing arithmetic progressions, Scripta Mathematica, 29 (1973): 1724. [Annotated copy of offprint]
Index entries for linear recurrences with constant coefficients, signature (0,2,0,1).


FORMULA

a(n) = A026741(3*n+1).
a(n)*A026741(n) = A005449(n).
a(n)*A022998(n+1) = A000567(n+1).
a(n) = A026741(n+1) + A022998(n).
a(2n) = A016921(n). a(2n+1) = A016789(n).
a(2n+1)*A026741(2n) = A045944(n).
G.f.: (1+2*x+5*x^2+x^3)/((x1)^2 * (1+x)^2).  R. J. Mathar, Sep 26 2009
a(n) = (3+9*n)/4 + (1)^n*(1+3*n)/4.  R. J. Mathar, Sep 26 2009
a(n) = 2*(3n+1)/(4((2n+2) mod 4)).  Bill McEachen, Jan 09 2015
If a(2n1) = x then a(2n) = 2x+3.  Robert G. Wilson v, Jan 26 2015
Let the reduced Collatz procedure be defined as Cr(n) = (3*n+1)/2. For odd n, a(n) = Cr(n). For even n, a(n) = Cr(4*n+1)/2.  Joe Slater, Nov 29 2016
a(n) = A067745(n+1) * 2^A007814((3n+1)/2).  Joe Slater, Nov 30 2016
a(n) = 2*a(n2)  a(n4).  G. C. Greubel, Apr 13 2017


MATHEMATICA

f[n_] := If[ OddQ@ n, (3n +1)/2, (3n +1)]; Array[f, 66, 0] (* Robert G. Wilson v, Jan 26 2015 *)
f[n_] := (3 (1)^(2n) + (1)^(1 + n)) (2 + 3n)/4; Array[f, 66] (* or *)
CoefficientList[ Series[(x^3 + 5x^2 + 2x + 1)/(x^2  1)^2, {x, 0, 65}], x] (* or *)
LinearRecurrence[{0, 2, 0, 1}, {1, 2, 7, 5}, 66] (* Robert G. Wilson v, Apr 13 2017 *)


PROG

(PARI) a(n)=n+=2*n+1; if(n%2, n, n/2) \\ Charles R Greathouse IV, Jan 13 2015


CROSSREFS

Cf. A165351, A165367, A067745, A007814.
Sequence in context: A135076 A155546 A073132 * A038004 A100792 A096037
Adjacent sequences: A165352 A165353 A165354 * A165356 A165357 A165358


KEYWORD

nonn,easy


AUTHOR

Paul Curtz, Sep 16 2009


EXTENSIONS

All comments changed to formulas by R. J. Mathar, Sep 26 2009
New name from Charles R Greathouse IV, Jan 13 2015
Name corrected by Joe Slater, Nov 29 2016


STATUS

approved



