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A319996 Let g = A006530(n), the largest prime factor of n. This filter sequence combines (g mod 6), n/g (A052126), and a single bit A319988(n) telling whether the largest prime factor is unitary.

%I A319996 #14 Jan 19 2019 04:15:43
%S A319996 1,2,3,4,5,6,7,8,9,10,5,11,7,12,13,14,5,15,7,16,17,10,5,18,19,12,20,
%T A319996 21,5,22,7,23,13,10,24,25,7,12,17,26,5,27,7,16,28,10,5,29,30,31,13,21,
%U A319996 5,32,33,34,17,10,5,35,7,12,36,37,24,22,7,16,13,38,5,39,7,12,40,21,41,27,7,42,43,10,5,44,33,12,13,26,5,45,46,16,17,10,24,47,7,48,28,49,5,22,7,34
%N A319996 Let g = A006530(n), the largest prime factor of n. This filter sequence combines (g mod 6), n/g (A052126), and a single bit A319988(n) telling whether the largest prime factor is unitary.
%C A319996 Restricted growth sequence transform of triple [A010875(A006530(n)), A052126(n), A319988(n)], with a separate value allotted for a(1).
%C A319996 Many of the same comments as given in A319717 apply also here, except for this filter, the "blind spot" area (where only unique values are possible for a(n)) is different, and contains at least all numbers in A070003. Because presence of 2 or 3 in the prime factorization of n do not force the value of a(n) unique, this is substantially less lax (i.e., more exact) filter than A319717. Here among the first 100000 terms, only 2393 have a unique value, compared to 74355 in A319717.
%C A319996 For all i, j:
%C A319996   a(i) = a(j) => A002324(i) = A002324(j),
%C A319996   a(i) = a(j) => A067029(i) = A067029(j),
%C A319996   a(i) = a(j) => A071178(i) = A071178(j),
%C A319996   a(i) = a(j) => A077462(i) = A077462(j) => A101296(i) = A101296(j),
%C A319996   a(i) = a(j) => A319690(i) = A319690(j).
%H A319996 Antti Karttunen, <a href="/A319996/b319996.txt">Table of n, a(n) for n = 1..100000</a>
%e A319996 For n = 15 (3*5) and n = 33 (3*11), the mod 6 residue of the largest prime factor is 5, also in both cases it is unitary (A319988(n) = 1), and the quotient n/A006530(n) is equal, in this case 3. Thus a(15) and a(33) are alloted the same running count (13 in this case) by rgs-transform.
%e A319996 For n = 2275 (5^2 * 7 * 13), n = 3325 (5^2 * 7 * 19), 5425 (5^2 * 7 * 31) and 6475 (5^2 * 7 * 37), the largest prime factor = 1 (mod 6), and A052126(n) = 175, thus these numbers are allotted the same running count (394 in this case) by rgs-transform.
%o A319996 (PARI)
%o A319996 up_to = 100000;
%o A319996 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 A319996 A006530(n) = if(n>1, vecmax(factor(n)[, 1]), 1);
%o A319996 A052126(n) = (n/A006530(n));
%o A319996 A319988(n) = ((n>1)&&(factor(n)[omega(n),2]>1));
%o A319996 A319996aux(n) = if(1==n,0,[A006530(n)%6, A052126(n), A319988(n)]);
%o A319996 v319996 = rgs_transform(vector(up_to,n,A319996aux(n)));
%o A319996 A319996(n) = v319996[n];
%Y A319996 Cf. A006530, A052126, A319988, A319717.
%Y A319996 Cf. A007528 (positions of 5's), A002476 (of 7's), A112774 (after its initial term gives the position of 10's in this sequence).
%Y A319996 Cf. also A319994 (modulo 4 analog).
%K A319996 nonn
%O A319996 1,2
%A A319996 _Antti Karttunen_, Oct 05 2018