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%I A351454 #12 Feb 14 2022 11:21:58
%S A351454 1,2,3,2,4,3,5,2,6,4,7,3,8,5,9,2,10,6,11,4,12,7,13,3,14,8,15,5,16,9,
%T A351454 17,2,18,10,19,6,20,11,21,4,22,12,23,7,24,13,25,3,26,14,27,8,28,15,29,
%U A351454 5,30,16,31,9,32,17,33,2,34,18,35,10,36,19,37,6,38,20,39,11,40,21,41,4,42,22,43,12,44,23,45,7,46,24,47,13,48,25,49,3,50,26,51,14,52,27,53,8,54
%N A351454 Lexicographically earliest infinite sequence such that a(i) = a(j) => A006530(i) = A006530(j), A329697(i) = A329697(j) and A331410(i) = A331410(j) for all i, j >= 1.
%C A351454 Restricted growth sequence transform of the triplet [A006530(n), A329697(n), A331410(n)], or equally, of the ordered pair [A006530(n), A335880(n)].
%C A351454 For all i, j >= 1: A324400(i) = A324400(j) => a(i) = a(j).
%H A351454 Antti Karttunen, <a href="/A351454/b351454.txt">Table of n, a(n) for n = 1..65537</a>
%e A351454 a(99) = a(121) because 99 = 3^2 * 11 and 121 = 11^2, so they have equal largest prime factor (A006530), and they also agree on A329697(99) = A329697(121) = 4 and on A331410(99) = A331410(121) = 4, therefore they get equal value (which is 51) allotted to them by the restricted growth sequence transform. - _Antti Karttunen_, Feb 14 2022
%o A351454 (PARI)
%o A351454 up_to = 65537;
%o A351454 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; };
%o A351454 A006530(n) = if(1==n, n, my(f=factor(n)); f[#f~, 1]);
%o A351454 A329697(n) = if(!bitand(n,n-1),0,1+A329697(n-(n/vecmax(factor(n)[, 1]))));
%o A351454 A331410(n) = if(!bitand(n,n-1),0,1+A331410(n+(n/vecmax(factor(n)[, 1]))));
%o A351454 Aux351454(n) = [A006530(n), A329697(n), A331410(n)];
%o A351454 v351454 = rgs_transform(vector(up_to, n, Aux351454(n)));
%o A351454 A351454(n) = v351454[n];
%Y A351454 Cf. A006530, A329697, A331410, A335880.
%Y A351454 Cf. also A324400, A336936, A351453.
%Y A351454 Differs from A351452 for the first time at n=49, where a(49) = 26, while A351452(49) = 19.
%Y A351454 Differs from A351460 for the first time at n=121, where a(121) = 51, while A351460(121) = 62.
%Y A351454 Differs from A103391(1+n) for the first time after n=1 at n=121, where a(121) = 51, while A103391(122) = 62.
%K A351454 nonn
%O A351454 1,2
%A A351454 _Antti Karttunen_, Feb 11 2022

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