jaja, I tried a brute force method using Mathematica...lol I give it the list of values and it gave out instantly:
(1/9) * (57 + 48 * n + 15 * Cos((2 * n * Pi)/3) + Sqrt(3) * Sin((2 * n * Pi)/3) )
Added by Jesus Galvez at 7:42am on November 27, 2012
triangulate my mesh faces thus i was getting 7 points per polygon. 7 outputs needed. It works with 7 outputs even if 3 of them are not being used.
Problem solved!
Thanks so much for your help!
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now I want to combine some branches together ,the rule is : For path{2} contain number 2 and 5, then conbine the two paths together ,and for path{5} includes only 2&5,no other number ,so it's end .For path{3}, includes number 3&6 ,so we go to path{6}, path{6} includes 3&6&18, then wo go to path{18} , path{18} contains a new number 27, so we check path{27} ,path{27} includes only 27&18, no new numbers ,so it is end.
With this logic, path{2}&{5} become one tree finally , the contains is 2&5 ,and so path{3}&{6} &{18} &{27}(the contents is 3,6,18,27), and so others .
so what I want is:
{2}(2,5)+{5}(2,5)={2/5/anything}(2,5) ## the new path index doesnot matter{3}(3,6)+{6}(3,6,18)+{18}(18,27)+{27}(27,18)={3/6/18/27/?}(3,6,18,27) ``````etc
I tried path mapper, but I donot think it can do the trick this time. may be I just miss something very visible?? Awaiting for your kind help~Thanks in advance.…
dont get you, i am saying sleect numbers in range 1 to 10, starting from 1 with a step of 2.
1 to 10 by 3 = 1 4 7 10
1 to 10 by 5 = 1 6
1 to 10 by 1 = 1 to 10 = 1 2 3 4 5 6 7 8 9 10
Added by Steve Lewis at 3:15pm on November 11, 2013