%I
%S 1,2,3,4,5,6,7,8,9,10,6,11,12,13,13,14,10,15,16,17,18,19,13,20,14,17,
%T 21,22,23,24,18,25,26,27,26,28,29,20,22,30,17,31,32,33,34,24,26,35,36,
%U 37,24,38,27,39,24,39,40,30,20,41,42,31,43,44,33,45,46,47,31,45,24,48,49,50,35,51,31,41,40,52,53,47,33,54,55,56,39,57,30,58,59,41,60
%N Restricted growth sequence computed for sigma, A000203.
%C When filtering sequences (by equivalence class partitioning), this sequence can be used instead of A000203, because for all i, j it holds that: a(i) = a(j) <=> A000203(i) = A000203(j) <=> A286358(i) = A286358(j).
%C Note that the latter equivalence indicates that this is also the restricted growth sequence of A286358.
%H Antti Karttunen, <a href="/A286603/b286603.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Si#SIGMAN">Index entries for sequences related to sigma(n)</a>
%e Construction: we start with a(1)=1 for sigma(1)=1 (where sigma = A000203), and then after, for all n > 1, whenever the value of sigma(n) has not been encountered before, we set a(n) to the least natural number k not already in sequence among a(1) .. a(n1), otherwise [whenever sigma(n) = sigma(m), for some m < n], we set a(n) = a(m), i.e., to the same value that was assigned to a(m).
%e For n=2, sigma(2) = 3, not encountered before, thus we allot for a(2) the least so far unused number, which is 2, thus a(2) = 2.
%e For n=3, sigma(3) = 4, not encountered before, thus we allot for a(3) the least so far unused number, which is 3, thus a(3) = 3.
%e For n=4, sigma(4) = 7, not encountered before, thus we allot for a(4) the least so far unused number, which is 4, thus a(4) = 4.
%e For n=5, sigma(5) = 6, not encountered before, thus we allot for a(5) the least so far unused number, which is 5, thus a(5) = 5.
%e For n=6, sigma(6) = 12, not encountered before, thus we allot for a(6) the least so far unused number, which is 6, thus a(6) = 6.
%e And this continues for n=7..10 because also for those n sigma obtains fresh new values, so here a(n) = n up to n = 10.
%e But then comes n=11, where sigma(11) = 12, a value which was already encountered at n=6 for the first time, thus we set a(11) = a(6) = 6.
%t With[{nn = 93}, Function[s, Table[Position[Keys@ s, k_ /; MemberQ[k, n]][[1, 1]], {n, nn}]]@ Map[#1 > #2 & @@ # &, Transpose@ {Values@ #, Keys@ #}] &@ PositionIndex@ Array[DivisorSigma[1, #] &, nn]] (* _Michael De Vlieger_, May 12 2017, Version 10 *)
%o (PARI)
%o A000203(n) = sigma(n);
%o 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; };
%o write_to_bfile(start_offset,vec,bfilename) = { for(n=1, length(vec), write(bfilename, (n+start_offset)1, " ", vec[n])); }
%o write_to_bfile(1,rgs_transform(vector(10000,n,A000203(n))),"b286603.txt");
%Y Cf. A000203, A206036, A211656, A286358.
%Y Cf. also A101296, A286605, A286610, A286619, A286621, A286622, A286626, A286378.
%K nonn
%O 1,2
%A _Antti Karttunen_, May 11 2017
