

A087445


Numbers that are congruent to 1 or 5 mod 12.


5



1, 5, 13, 17, 25, 29, 37, 41, 49, 53, 61, 65, 73, 77, 85, 89, 97, 101, 109, 113, 121, 125, 133, 137, 145, 149, 157, 161, 169, 173, 181, 185, 193, 197, 205, 209, 217, 221, 229, 233, 241, 245, 253, 257, 265, 269, 277, 281, 289, 293, 301, 305, 313, 317, 325, 329
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

From Bob Selcoe, Jun 03 2015: (Start)
For k >= 1: all numbers congruent to A002450(k) mod 2^(2k+1) and A072197(k) mod 4^(k+1) not congruent to 0 mod 3. Equivalently, for k >= 3: all numbers congruent to A096773(k) mod 2^k not congruent to 0 mod 3.
Conjecture: at least one number in this sequence must appear in all Collatz sequences.
(End)
The sequence is composed of all numbers in congruence classes T(n,1) mod 2^(n+k) in A259663 (i.e., T"(1) in array T259663(n,k)) not congruent to 0 mod 3. Therefore the conjecture above is true (see A259663 for additional explanation).  Bob Selcoe, Jul 15 2017
Closure of {1} under the map (x,y)>2x+3y [KlarnerRado, see Lagarias (2016), p. 755].  N. J. A. Sloane, Oct 06 2016
The above conjecture is true: this is because even numbers and odd numbers divisible by 3 will lead to the set of odd numbers not divisible by 3. Odd numbers of the form 4k  1 can also be ignored, as this consists of odd numbers that grow between themselves and the next odd term through Collatz iteration. No infinite sequence of growth between consecutive odd terms is possible, so all numbers of the form 4k  1 will lead to an odd number that shrinks between itself and the next odd number. All numbers 4k  1 will lead to a number in 4k  3, the odd numbers that shrink between themselves and the following odd term. What we are left after that elimination is this sequence.  Aidan Simmons, Feb 25 2019


LINKS

Table of n, a(n) for n=1..56.
J. C. Lagarias, ErdÅ‘s, Klarner and the 3x+ 1 Problem, Amer. Math. Monthly 123 (No. 8, 2016), 753776.
Index entries for linear recurrences with constant coefficients, signature (1,1,1).


FORMULA

G.f.: x*(1+4*x+7*x^2)/((1+x)*(1x)^2).
E.g.f.: 6*(x1)*exp(x) + 7  exp(x).  corrected by Robert Israel, Jun 10 2015
a(n) = 6*(n1)  (1)^n.  Rolf Pleisch, Aug 04 2009
a(n) = 12*n  a(n1)  18 (with a(1)=1).  Vincenzo Librandi, Aug 08 2010
a(n) = a(n1) + a(n2)  a(n3), with a(0)=1, a(1)=5, a(2)=13.  Harvey P. Dale, Jun 13 2011


MAPLE

seq(6*(n1)(1)^n, n=1..100); # Robert Israel, Jun 10 2015


MATHEMATICA

LinearRecurrence[{1, 1, 1}, {1, 5, 13}, 70] (* or *) Rest[CoefficientList[ Series[x (1+4x+7x^2)/((1+x)(1x)^2), {x, 0, 70}], x]] (* Harvey P. Dale, Jun 13 2011 *)


PROG

(PARI) a(n)=(n1)\2*12 + [5, 1][n%2+1] \\ Charles R Greathouse IV, Jun 03 2015
(MAGMA) [k:k in [1..330] k mod 12 in [1, 5]]; // Marius A. Burtea, Feb 08 2020


CROSSREFS

Cf. A001651, A047241, A087444, A087446.
Cf. A259663.
Sequence in context: A105596 A037046 A126887 * A020882 A081804 A004613
Adjacent sequences: A087442 A087443 A087444 * A087446 A087447 A087448


KEYWORD

easy,nonn


AUTHOR

Paul Barry, Sep 04 2003


STATUS

approved



