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A319717 Filter sequence combining the largest proper divisor of n (A032742) with modulo 6 residue of the smallest prime factor, A010875(A020639(n)), and a single bit A319710(n) telling whether the smallest prime factor is unitary.

%I A319717 #34 Jul 15 2022 06:48:00
%S A319717 1,2,3,4,5,6,7,8,9,10,5,11,7,12,13,14,5,15,7,16,17,18,5,19,20,21,22,
%T A319717 23,5,24,7,25,26,27,28,29,7,30,31,32,5,33,7,34,35,36,5,37,38,39,40,41,
%U A319717 5,42,43,44,45,46,5,47,7,48,49,50,51,52,7,53,54,55,5,56,7,57,58,59,60,61,7,62,63,64,5,65,66,67,68,69,5,70,71,72,73,74,75,76,7,77,78,79,5,80,7,81,82,83,5,84,7,85,86,87,5,88,89,90,91,92,93,94,95
%N A319717 Filter sequence combining the largest proper divisor of n (A032742) with modulo 6 residue of the smallest prime factor, A010875(A020639(n)), and a single bit A319710(n) telling whether the smallest prime factor is unitary.
%C A319717 Restricted growth sequence transform of triple [A010875(A020639(n)), A032742(n), A319710(n)] (with a separate value allotted for a(1)), or equally, of ordered pair [A319716(n), A319710(n)].
%C A319717 In addition to A319716, this filter sequence also records in the value of a(n) also the fact whether the smallest prime factor of n is unitary or not. This information is enough to determine the modulo 6 residues of all the divisors of n, thus sequences like A002324 are essentially functions of this sequence. Moreover, a lot of other information is immediately (and unavoidably) present, for example the exact prime signature of n, including also the relative order of exponents.
%C A319717 Any such filtering sequence can be perceived also in terms of what information it leaves out from a(n) that would be needed to reconstruct whole n from each a(n). If the whole n could be reconstructed from a(n) each time, then sequence a would be injective, and would be useless for filtering, because then it would match with any sequence. In this filter, what is left out is only the exact identity of the smallest prime factor, although its residue class mod 6 is retained. However, when the smallest prime factor is 2 or 3, this can be seen from that residue value, so for any number x in A047229, both A020639(x) and A032742(x) are known, and as x = A020639(x)*A032742(x), it means such numbers must occur in their own singleton equivalence classes.
%C A319717 Likewise, for any n in A283050, even if not divisible by 2 or 3, when we have A319710(n) stored in the triple as 1, this immediately gives away the exact identity of the smallest prime factor, which is equal to A014673(n) = A020639(A032742(n)) in these cases.
%C A319717 Thus there is a substantial subset of N (containing at least the union of A047229 and A283050) which is actually in the "blind sector" of this filter, "where anything goes", as this sequence obtains only unique values in that subdomain.
%C A319717 There is a related filter sequence A319996, which operates by "cleaving n from its high end" (by storing the residue class of the largest prime factor, A006530, instead of the smallest, together with n/A006530(n)), which has its own blind spots, but fortunately, they do not fully coincide with the blind spots of this filter. Naturally, any sequence like A002324 should match both to this sequence and A319996.
%C A319717 For all i, j:
%C A319717   a(i) = a(j) => A002324(i) = A002324(j),
%C A319717   a(i) = a(j) => A067029(i) = A067029(j),
%C A319717   a(i) = a(j) => A071178(i) = A071178(j),
%C A319717   a(i) = a(j) => A077462(i) = A077462(j) => A101296(i) = A101296(j),
%C A319717   a(i) = a(j) => A319716(i) = A319716(j) => A319690(i) = A319690(j).
%H A319717 Antti Karttunen, <a href="/A319717/b319717.txt">Table of n, a(n) for n = 1..100000</a>
%e A319717 For n = 65 = 5*13 and 143 = 11*13, the smallest prime factor is of the form 6k+5,  doesn't occur more than once in the factorization, and the largest proper divisor is the same number (13) in both cases, thus a(65) = a(143) (= 51, a running count value allotted by rgs-transform for this equivalence class).
%e A319717 For n = 1805 (5*19^2), 3971 (11*19^2), 6137 (17*19^2), it's like above, but the largest proper divisor is in all three cases 361 = 19^2, thus a(1805) = a(3971) = a(6137) (= 1405).
%e A319717 Note that such nontrivial equivalence classes may only contain numbers that are 5-rough, A007310, with no prime factors 2 or 3, and also, they may not contain numbers from A283050. See the comments section.
%o A319717 (PARI)
%o A319717 up_to = 100000;
%o A319717 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 A319717 A032742(n) = if(1==n,n,n/vecmin(factor(n)[,1]));
%o A319717 A286476(n) = if(1==n,n,(6*A032742(n) + (n % 6)));
%o A319717 A319710(n) = ((n>1)&&(factor(n)[1,2]>1));
%o A319717 v319717 = rgs_transform(vector(up_to,n,[A286476(n),A319710(n)]));
%o A319717 A319717(n) = v319717[n];
%Y A319717 Cf. A101296, A319707, A319716, A319996.
%Y A319717 Cf. also A320004 (analogous sequence for modulo 4 residues).
%Y A319717 Differs from A319707 for the first time at n=143, where a(143) = 51, differs from A319716 for the first time at n=121, where a(121) = 95.
%K A319717 nonn
%O A319717 1,2
%A A319717 _Antti Karttunen_, Oct 04 2018