Guest Posted April 19, 2009 Report Share Posted April 19, 2009 1,2,3,2,5,1,7,2,3,1,11,3,13,1,5,2,17,1,?,?,?,?,?,? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 19, 2009 Report Share Posted April 19, 2009 (edited) Romulus064 said: 1,2,3,2,5,1,7,2,3,1,11,3,13,1,5,2,17,1,?,?,?,?,?,? Reveal hidden contents the nth term of the sequence is the smallest factor of n which can be multiplied by some subgroup of terms 1 through n-1 of the sequence to produce n? So the next 6 terms would be: 19,1,1,1,23,1 EDIT: Nevermind, this doesn't work for the 12th or 15th terms. Edited April 19, 2009 by hookemhorns Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 19, 2009 Report Share Posted April 19, 2009 Romulus064 said: 1,2,3,2,5,1,7,2,3,1,11,3,13,1,5,2,17,1,?,?,?,?,?,? is the answer 3, 19, 2, 7, 2, 23? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 19, 2009 Report Share Posted April 19, 2009 (edited) Reveal hidden contents 19,5,1,11,23,1 Reveal hidden contents 19,1,7,11,23,1 Romulus064 said: 1,2,3,2,5,1,7,2,3,1,11,3,13,1,5,2,17,1,?,?,?,?,?,? Edited April 19, 2009 by varnejm Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 19, 2009 Report Share Posted April 19, 2009 Romulus064 said: 1,2,3,2,5,1,7,2,3,1,11,3,13,1,5,2,17,1,?,?,?,?,?,? Reveal hidden contents 19 5 7 11 23 n= highest factor that has already been in the sequence....(i hope) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 19, 2009 Report Share Posted April 19, 2009 I can see a pattern of course... the primes are easily recognizable Reveal hidden contents I don't see the real sequence. a(n) = 1 , except when: - a is a prime, or - a is a prime power of a(n) If this is even valid as a number sequence solution (given the exceptions) it fits most of the sequence. However, it does not work for n=12, or n=15. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 19, 2009 Report Share Posted April 19, 2009 uhre said: I can see a pattern of course... the primes are easily recognizable Reveal hidden contents I don't see the real sequence. a(n) = 1 , except when: - a is a prime, or - a is a prime power of a(n) If this is even valid as a number sequence solution (given the exceptions) it fits most of the sequence. However, it does not work for n=12, or n=15. Reveal hidden contents a(n) = 1 , except when: - n is a prime, or - n is a prime power of a(n) Still I did not get any closer to a solution. Any ideas? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted April 19, 2009 Report Share Posted April 19, 2009 (edited) varnejm said: Reveal hidden contents 19,1,7,11,23,1 Reveal hidden contents You're right! What I want to know is HTH you got that. Edited April 19, 2009 by Romulus064 Quote Link to comment Share on other sites More sharing options...
0 Pickett Posted April 20, 2009 Report Share Posted April 20, 2009 Romulus064 said: Reveal hidden contents You're right! What I want to know is HTH you got that. Reveal hidden contents I can get the same answer as varnejm, using this formula: f(n) = 1 if n has 2 distinct prime factors, one of which is a 2 or n = 1 f(n) = 2 if n is a power of 2 f(n) = n if n is prime f(n) = largest prime factor of n for all other cases However, the only number that doesn't seem to work with that is n=12..because the factorization of 12 is 2, 2, 3...which has 2 distinct prime factors, one of which is a 2...so, I would think that it would be "1"...but the sequence says it is "3"...so I think I'm missing something...but that's what I have so far. Quote Link to comment Share on other sites More sharing options...
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1,2,3,2,5,1,7,2,3,1,11,3,13,1,5,2,17,1,?,?,?,?,?,?
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