

A319716


Filter sequence combining the largest proper divisor of n (A032742) with modulo 6 residue of the smallest prime factor, A010875(A020639(n)).


4



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, 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, 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, 43
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OFFSET

1,2


COMMENTS

Restricted growth sequence transform of A286475, or equally, of A286476.
In each a(n) there is enough information to determine the modulo 6 residues of all the prime factors of n (when counted with multiplicity), thus sequences like A319690 and A319691 (which is the characteristic function of A004611) are essentially functions of this sequence. However, to determine that for all divisors of n, more information is needed. See A319717.
For all i, j:


LINKS



EXAMPLE

For n = 55 = 5*11 and 121 = 11*11, 55 = 121 = 1 mod 6 and 11 is their common largest proper divisor, thus they are allotted the same number by the restricted growth sequence transform, that is a(55) = a(121) = 43 (which is the number allotted). Note that such nontrivial equivalence classes may only contain numbers that are 5rough, A007310, with no prime factors 2 or 3.


PROG

(PARI)
up_to = 100000;
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; };
A032742(n) = if(1==n, n, n/vecmin(factor(n)[, 1]));
v319716 = rgs_transform(vector(up_to, n, A286476(n)));


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



