Aug 2, 2010 #1 smslca Member level 1 Joined Sep 9, 2007 Messages 33 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,286 Activity points 1,656 please convert the below(P) into formula using (m), or give an algorithm using (m) to get (P) m=2, P=0 m=3, P=3 m=4, P=3+5 m=5, P=3+5+5 m=6, P=3+5+5+7 m=7, P=3+5+5+7+9 m=8, P=3+5+5+7+9+9 m=9, P=3+5+5+7+9+9+11 m=10, P=3+5+5+7+9+9+11+13 m=11, P=3+5+5+7+9+9+11+13+13 m=12, P=3+5+5+7+9+9+11+13+13+15 m=13, P=3+5+5+7+9+9+11+13+13+15+17 m=14, P=3+5+5+7+9+9+11+13+13+15+17+17 and so on I did not understand how to evaluate it into a formula
please convert the below(P) into formula using (m), or give an algorithm using (m) to get (P) m=2, P=0 m=3, P=3 m=4, P=3+5 m=5, P=3+5+5 m=6, P=3+5+5+7 m=7, P=3+5+5+7+9 m=8, P=3+5+5+7+9+9 m=9, P=3+5+5+7+9+9+11 m=10, P=3+5+5+7+9+9+11+13 m=11, P=3+5+5+7+9+9+11+13+13 m=12, P=3+5+5+7+9+9+11+13+13+15 m=13, P=3+5+5+7+9+9+11+13+13+15+17 m=14, P=3+5+5+7+9+9+11+13+13+15+17+17 and so on I did not understand how to evaluate it into a formula
Aug 2, 2010 #2 _Eduardo_ Full Member level 5 Joined Aug 31, 2009 Messages 295 Helped 118 Reputation 238 Reaction score 103 Trophy points 1,323 Location Argentina Activity points 2,909 Re: convert into formula Let n = [m/3] ; [] mean integer part and k = m mod 3 ; mod mean modulus (remainder of the integer divide m/3) That is: m --> n , k 3 --> 1, 0 4 --> 1, 1 5 --> 1, 2 6 --> 2, 0 7 --> 2, 1 8 --> 2, 2 9 --> 3, 0 ............... Then P(n,k) = n*(6*n-1) + k*(4*n+1) - 2
Re: convert into formula Let n = [m/3] ; [] mean integer part and k = m mod 3 ; mod mean modulus (remainder of the integer divide m/3) That is: m --> n , k 3 --> 1, 0 4 --> 1, 1 5 --> 1, 2 6 --> 2, 0 7 --> 2, 1 8 --> 2, 2 9 --> 3, 0 ............... Then P(n,k) = n*(6*n-1) + k*(4*n+1) - 2
Aug 3, 2010 #3 smslca Member level 1 Joined Sep 9, 2007 Messages 33 Helped 0 Reputation 0 Reaction score 0 Trophy points 1,286 Activity points 1,656 convert into formula thanks for the replies