A096773 a(n+2) = 4*a(n) + 1; a(1) = 0, a(2) = 3.

0, 3, 1, 13, 5, 53, 21, 213, 85, 853, 341, 3413, 1365, 13653, 5461, 54613, 21845, 218453, 87381, 873813, 349525, 3495253, 1398101, 13981013, 5592405, 55924053, 22369621, 223696213, 89478485, 894784853, 357913941, 3579139413, 1431655765

Offset

1,2

Comments

Remainders for classes m of integers n (mod 2^(m+1)). After applying one Collatz (3x+1)-transformation to the so-classified integers the result can be written in two classes (mod 6) only.

This classifying scheme covers all positive integers.

With one 3x+1-transformation T(x;p) := x' = (3x+1)/2^p, all numbers x, described in the form, with the free parameter i >= 0, x = i*2^N + a(N) result in x', describable by the two classes with the same parameter i:

x' = i*6 + 1 (for odd N>2), or x' = i*6 + 5 (for even N). Thus

x = 4*i + 3 -> x' = 6*i + 5, x = 8*i + 1 -> x' = 6*i + 1,

x = 16*i + 13 -> x' = 6*i + 5, x = 32*i + 5 -> x' = 6*i + 1,

x = 64*i + 53 -> x' = 6*i + 5, x = 128*i + 21 -> x' = 6*i + 1,

....

all with "i" as a free parameter >= 0 covering all positive integers.

Links

Michael De Vlieger, Table of n, a(n) for n = 1..3323

Index entries for sequences related to 3x+1 (or Collatz) problem

Index entries for linear recurrences with constant coefficients, signature (1,4,-4).

Formula

a(2m) = (5*2^(2m-1) - 1)/3, a(2m-1) = (2^(2m-2)-1)/3.

From Paul Curtz, Jul 01 2008; corrected by Bob Selcoe, Jul 28 2018: (Start)

a(2n) = 10*a(2n-1) + 3.

a(n+1) - 2*a(n) = A001045(n+2), signed. (End)

a(n) = (2^(n-1)*(3 + 2*(-1)^n) - 1)/3. - L. Edson Jeffery, Jul 12 2015

a(2n) = A086893(2n), a(2n+1) = A086893(2n-1), n > 0. - Yosu Yurramendi, Jan 17 2017

G.f.: -x^2*(-3+2*x) / ( (x-1)*(2*x+1)*(2*x-1) ). - R. J. Mathar, Mar 07 2017

a(2n) = A072197(n-1), n > 0; a(2n+1) = A002450(n), n >= 0. - Yosu Yurramendi, Mar 07 2017

a(2n) = (A266753(n) + A004171(n-1))/2, a(2n+1) = (A266753(n) - A004171(n-1))/2, n > 0. - Yosu Yurramendi, Mar 07 2017

a(n) = least residue 2*3^(2^(n-4)-1) - 1 (mod 2^n), n >= 5. - Bob Selcoe, Jul 26 2018

Example

a(1) = (2^0-1)/3 =  0, a(2) = (5*2^1 - 1) / 3 =  3,
a(3) = (2^2-1)/3 =  1, a(4) = (5*2^3 - 1) / 3 = 13,
a(5) = (2^4-1)/3 =  5, a(6) = (5*2^5 - 1) / 3 = 53,
a(7) = (2^6-1)/3 = 21.
....

Mathematica

a[1] = 0; a[2] = 3; a[n_] := a[n] = 4a[n - 2] + 1; Table[ a[n], {n, 35}] (* Robert G. Wilson v, Aug 20 2004 *)
Table[(2^(n - 1)*(3 + 2*(-1)^(n)) - 1)/3, {n, 10}] (* L. Edson Jeffery, Jul 12 2015 *)

Prog

(Magma) [(2^(n-1)*(3 + 2*(-1)^n) - 1)/3: n in [1..40]]; // Vincenzo Librandi, Jul 12 2015
(Perl) # To map any (odd) v to its (r, c):
use bigint; $v=149; $r=$c=0; while(1){ $b=($v&1); $v>>=1; if ($b==($v&1)){ $c=($v>>1); last} $r++} $r&=1; # this splits the binary representation into two parts, at the first repeated digit from the right: the number of bits on the right is the row value, and the binary value on the left is the column value. Example: 149 => 1.00.10101 => (r, c)=(5, 1). Ruud H.G. van Tol, Sep 23 2021

Crossrefs

Bisections are A002450 & A072197.

After the initial 0, column 1 of A257852.

Sequence in context: A113139 A266577 A143411 * A118384 A341725 A258239

Adjacent sequences: A096770 A096771 A096772 * A096774 A096775 A096776

Keyword

easy,nonn,changed

Author

Gottfried Helms, Aug 15 2004

Status

approved