OFFSET
1,3
COMMENTS
The map takes a positive odd integer x (= 2*n-1) and produces the positive odd integer a(n).
Farkas proves that the trajectory of the iterates of the map starting from any positive odd integer always reaches 1.
REFERENCES
H. M. Farkas, "Variants of the 3N+1 Conjecture and Multiplicative Semigroups", in Entov, Pinchover and Sageev, "Geometry, Spectral Theory, Groups, and Dynamics", Contemporary Mathematics, vol. 387, American Mathematical Society, 2005, p. 121.
J. C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, American Mathematical Society, 2010, p. 74.
LINKS
Paolo Xausa, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,2,0,0,0,0,0,-1).
EXAMPLE
From Omar E. Pol, Jan 01 2022: (Start)
Written as a rectangular array with six columns read by rows the sequence begins:
1, 1, 3, 11, 3, 17;
7, 5, 9, 29, 7, 35;
13, 9, 15, 47, 11, 53;
19, 13, 21, 65, 15, 71;
25, 17, 27, 83, 19, 89;
31, 21, 33, 101, 23, 107;
37, 25, 39, 119, 27, 125;
43, 29, 45, 137, 31, 143;
49, 33, 51, 155, 35, 161;
55, 37, 57, 173, 39, 179;
...
(End)
MATHEMATICA
LinearRecurrence[{0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, -1}, {1, 1, 3, 11, 3, 17, 7, 5, 9, 29, 7, 35}, 100]
Table[Which[Mod[n, 3]==0, n/3, Mod[n, 4]==3, (3n+1)/2, True, (n+1)/2], {n, 1, 200, 2}] (* Harvey P. Dale, May 15 2022 *)
PROG
(PARI) a(n)=if (n%3==2, 2*n\3, if (n%2, n, 3*n-1)) \\ Charles R Greathouse IV, Nov 16 2021
(Python)
def a(n):
x = 2*n - 1
return x//3 if x%3 == 0 else ((3*x+1)//2 if x%4 == 3 else (x+1)//2)
print([a(n) for n in range(1, 66)]) # Michael S. Branicky, Nov 16 2021
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paolo Xausa, Nov 16 2021
STATUS
approved