As well remarked by Franklin, (a) and (b) can be combined in a single
function with negative integers in the sequences.
However... at least because my program ran as I defined previously... with
the initial definitions (a) and (b) using positive integers,
in the range of odd integers from 1 to 10,000,000,000:
1) All integers fall in
(a) 1 or 17 families
(b) 1 or 13 families
2) Longest ways to get 1:
(a) 212 steps with 8,316,337,203
(b) 206 steps with 6,337,579,165
3) Longest ways to get 17(a) or 13(b):
(a) 190 steps with 7,575,090,961
(b) 204 steps with 8,774,370,049
4) Percentage P1 of numbers giving 1, from 1 to n
(a) 78.45% < P1 < 79.15%
(b) same
5) Percentage P1' of numbers giving 1, from n to n+100,000,000
(a) 62% < P1' < 94%
(b) same
6) Biggest reached term within sequences
(a) 32,705,624,140,216,093,177
with n = 5,376,042,513
(and same term for some others bigger n)
(b) 24,529,218,105,162,069,883
with n = 4,032,031,885
Examples on 5)
>From 6,400,000,000 to 6,500,000,000:
P1'(a) = 93.1%
Rrom 6,800,000,000 to 6,900,000,000:
P1'(a) = 62.4%
Supplemental remarks on 4) and 5)
The oscillations of (a) and (b) are complementary (when P1'(a) is big,
P1'(b) is small). Periods increasing with n.
Maybe some other percentages or oscillations that I can't see, within n to
n+100,000,000.
Christian.
-----Message d'origine-----
De : Christian Boyer [mailto:cboyer@club-internet.fr]
Envoyé : dimanche 17 septembre 2006 22:49
À : 'math-fun'
Objet : RE: [math-fun] Collatz 3N+1 problems
After Dan's (a), I propose to define a second function (b):
(a) M=(3N+1)/2^K then P=(3M-1)/2^K.
(b) M=(3N-1)/2^K then P=(3N+1)/2^K.
Interesting to see very similar properties:
(a) two families, 1 and 17 (-> 19 -> 43 -> 97 -> 109 -> 61 -> 17...)
(b) two families, 1 and 13 (-> 29 -> 65 -> 73 -> 41 -> 23 -> 13...)
Similar percentages 1:17 for (a) and 1:13 for (b)???
Soon a report on the 2 functions from 1 to 10,000,000,000.
Christian.
