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Posted: Fri Aug 03, 2007 2:47 pm
by sonyckson
Thanks... no i did not find the solution por non primes... jeje i solved the problem but just finding the pattern (for primes... ).. and now i see you are right.. its number theory ... thank you. eric.
Posted: Sat Aug 11, 2007 10:49 am
by ibroker
I can't understand problem exactly.
Problem says, for p=3 the infinite sequence is 1,2,4,5,10,11,13,14...
Why is "1,2,4,5,7,8,10,11..." sequence not allowed?
I think that this sequence doesn't contain arithmetic sequence length p.
I need help.
Posted: Sat Aug 11, 2007 11:18 am
by hi!man!
The sequence 1,2,4,5,7,8,10,11,13,14,...
contain a lot of arithmetic sequence,
such as (1,4,7),(2,5,8 ),(1,7,13)...etc.
(1,4,7) --> 7-4=4-1=3 so it is arithmetic sequence.
Think more and you will get it
Posted: Sat Aug 11, 2007 1:56 pm
by ibroker
Oh! I see !
I think that subsequence always continuous.
So I consider only A
, A[i+1], A[i+2]... A[i+p-1](A[1]~A[n]),
not A, A[j], A[k] ... (i<j<k)
Thanks ! 