A286449 Restricted growth sequence computed for A033879 (deficiency), or equally, for A033880 (abundance of n).
1, 1, 2, 1, 3, 4, 5, 1, 6, 2, 7, 8, 9, 3, 5, 1, 10, 11, 12, 13, 7, 14, 15, 16, 17, 7, 18, 4, 19, 16, 20, 1, 12, 18, 15, 21, 22, 10, 15, 23, 24, 16, 25, 3, 9, 26, 27, 28, 29, 30, 20, 5, 31, 16, 32, 33, 34, 35, 36, 37, 38, 19, 15, 1, 27, 16, 39, 7, 25, 8, 40, 41, 42, 34, 35, 9, 36, 16, 43, 44, 29, 32, 45, 46, 47, 24, 48, 8, 49, 50, 40, 10, 36, 51, 40, 52, 53
Offset: 1
Keywords
Examples
We start by setting a(1) = 1 for A033879(1) = 1. Then, whenever A033879(k) is equal to some A033879(m) with m < k, we set a(k) = a(m). Otherwise (when the value is a new one, not encountered before), we allot for a(k) the least natural number not present among a(1) .. a(k-1). For n=2, as A033879(2) = 1, which was already present at A033879(1), we set a(2) = a(1) = 1. For n=3, as A033879(3) = 2, which is a new value not encountered before, we set a(3) = 1 + max(a(1),a(2)) = 2. For n=4, as A033879(4) = 1, which was already present at n = 2 and n = 1, we set a(4) = a(1) = 1. For n=5, as A033879(5) = 4, which is a new value not encountered before, we set a(5) = 1 + max(a(1),a(2),a(3),a(4)) = 3. For n=12, as A033879(12) = -4, which is a new value not encountered before, we set a(12) = 1 + max(a(1),...,a(11)) = 8. Note that the sign matters here; -4 is not equal to +4, which was encountered already at n=5.
Links
Programs
-
PARI
rgs_transform(invec) = { my(occurrences = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(occurrences,invec[i]), my(pp = mapget(occurrences, invec[i])); outvec[i] = outvec[pp] , mapput(occurrences,invec[i],i); outvec[i] = u; u++ )); outvec; }; write_to_bfile(start_offset,vec,bfilename) = { for(n=1, length(vec), write(bfilename, (n+start_offset)-1, " ", vec[n])); } A033879(n) = ((2*n)-sigma(n)); write_to_bfile(1,rgs_transform(vector(10000,n,A033879(n))),"b286449.txt");
Comments