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| • this is OUT: {{n, n, n, n}, {f1[n], f2[n], f3[n], f4[n]}, {f1[f1[n]], f2[f2[n]], f3[f3[n]], f4[f4[n]]}, {f1[f1[f1[n]]], f2[f2[f2[n]]], f3[f3[f3[n]]], f4[f4[f4[n]]]}} • at the 3rd step I want to get 16... |
| this output is the intersection of pairs of larger subsets: {{13, 24, 68, 79}, {13, 24, 68, 79, 90}, {13, 24, 44, 48, 68, 79}, {13, 20, 44, 48, 68, 79}, {13, 24, 44, 55, 68, 79, 90}, {13, 24, 44, 68, 79}, {13, 20, 24, 44, 48, 55, 79}, {13,... |
| I solved by myself in an inelegant way: I added these line
alpha={a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z}; incognite=alpha[[Range[26-cifre+2,26]]]; |
| It works for numbers 1 or 2 digits, but the reason why I want to get the output with 3-digit is because the output already contains a 3-digit numbers, with your formula the 3-digit numbers become 4-digit
there is a way to map placeholders, such... |
| tnk's i try |
| Prof. Gianluca, can i write to u in private @ university mail? |
| i fond "NORMAL" as great solution
Normal[Solve[{function}, {x,y}, Integers]] /. {C[1] -> 1, C[2] -> 1} remove "conditional expression" and replacing C[n] with 1 (if present as a product in solutions) u get a clear result. |
| conica=a x^2+b x y+c y^2+d x+e y+f==0; cinquepunti={{0,-3},{-1,4},{-7,6},{3,8},{4,8}} prima=conica/.{x->cinquepunti[[1,1]],y->cinquepunti[[1,2]]} seconda=conica/.{x->cinquepunti[[2,1]],y->cinquepunti[[2,2]]} ... |