A366286 Lexicographically earliest infinite sequence such that a(i) = a(j) => A366285(i) = A366285(j) for all i, j >= 0, where A366285(n) is the denominator of n / A366275(n).
1, 2, 2, 1, 2, 3, 1, 4, 2, 5, 3, 6, 1, 7, 4, 8, 2, 9, 5, 10, 3, 7, 6, 11, 1, 4, 7, 12, 4, 13, 8, 14, 2, 9, 9, 15, 5, 16, 10, 11, 3, 17, 7, 18, 6, 13, 11, 19, 1, 20, 4, 21, 7, 22, 12, 1, 4, 23, 13, 24, 8, 25, 14, 26, 2, 27, 9, 28, 9, 16, 15, 29, 5, 30, 16, 11, 10, 31, 11, 32, 3, 20, 17, 33, 7, 34, 18, 35, 6, 36, 13, 19
Offset: 0
Links
Crossrefs
Programs
-
PARI
up_to = 65537; rgs_transform(invec) = { my(om = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(om,invec[i]), my(pp = mapget(om, invec[i])); outvec[i] = outvec[pp] , mapput(om,invec[i],i); outvec[i] = u; u++ )); outvec; }; A366285(n) = { my(u=A366275(n)); (u/gcd(n,u)); }; \\ Uses also the program given in A366275. v366286 = rgs_transform(vector(1+up_to,n,A366285(n-1))); A366286(n) = v366286[1+n];
Comments