1. At least 1 statement among these 2n+1 are true.
2. At least 3 statements among these 2n+1 are false.
3. At least 5 statements among these 2n+1 are true.
...
2n. At least (4n1) statments among these 2n+1 are false.
2n+1. At least (4n+1) statements among these 2n+1 are true.
How many statements are true? Which?
(St = Statement; V = True; F = False) (n as in the sequence)
As I said in my first mail, St from n+1 to 2n + 1 are always F (they are impossible).
Then, some of the first even St will be V (and this also assures that the first St is necessarily V).
More: some odd and even St, placed just before the St number n, will always be F. Some odd St are F because the number of VSt has to be always inferior to n, and some of this St rule against it; some even St are also F, because the number of F St can't be so close to 2n+1, that there is no room for V St (some St are F because rule against this).
All of it explain the shape of each set of 2n+1 St. The shape is the same for the different sets, with some V at the beginning of the sequence, followed by V and F alternate and, at least, a lot of F terms. (Shape for n=13, for ex., is:
VVVVFVFVFVFFF FFFFFFFFFFFFFF (113 1427)
(And for n=7) : VVFVFVF FFFFFFFF (17 815)
I tried to obtain a general math formula, based on the process of obtention for V and F for each set, but I couldn't found an easy way. Anyway, I could noted that probably the sets of 2n+1 St, follows a 8cycle solution, depending on the value assigned to n. As I couldn't do it with maths, I tried directly to solve some of the 2n+1 sets of St, to get inducted formulas.
This are the results I get:
If n= 8k (=8, 16, 24, ....) there are two possible solutions(because the last assigned St admit both possibilities, V and F)
If n= 8k1 (=7, 15, 23...) V= (n+1)/2 F= (3n+1)/2
If n= 8k2 (=6, 14, 22...) V= n/2 F= (3n+2)/2
If n= 8k3 (=5, 13,...) (as for 8k1)
If n= 8k4 (= 4, 12, 20....) there is not a logical solution: assigning F to the last St convert the St automatically in V, and viceversa.
If n= 8k+1 (= 9, 17, ...) (as for 8k1)
If n= 8k+2 (= 10, 18,...) V= (n+2)/2 F= 3n/2
If n= 8k+3 (=11, 19,...) (as for 8k1)
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Examples: If n= 20 there is not a solution
If n = 37 (8*53) V= (37+1)/2 = 19 F= (3*37+1)/2 = 56
I've checked a little number of sequences anyway, so I'm not all sure of all this formulas.
Edited on June 14, 2005, 10:25 pm

Posted by armando
on 20050614 16:28:08 