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A328401
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Lexicographically earliest infinite sequence such that a(i) = a(j) => A328400(i) = A328400(j) for all i, j.
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2
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1, 2, 2, 3, 2, 2, 2, 4, 3, 2, 2, 5, 2, 2, 2, 6, 2, 5, 2, 5, 2, 2, 2, 7, 3, 2, 4, 5, 2, 2, 2, 8, 2, 2, 2, 3, 2, 2, 2, 7, 2, 2, 2, 5, 5, 2, 2, 9, 3, 5, 2, 5, 2, 7, 2, 7, 2, 2, 2, 5, 2, 2, 5, 10, 2, 2, 2, 5, 2, 2, 2, 11, 2, 2, 5, 5, 2, 2, 2, 9, 6, 2, 2, 5, 2, 2, 2, 7, 2, 5, 2, 5, 2, 2, 2, 12, 2, 5, 5, 3, 2, 2, 2, 7, 2
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OFFSET
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1,2
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COMMENTS
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For all i, j:
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LINKS
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EXAMPLE
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Numbers 2 (= 2^1), 3 (= 3^1), 6 = (2^1 * 3^1) and 30 (2^1 * 3^1 * 5^1) all have just one distinct exponent, 1, in the multisets of exponents that occur in their prime factorization, thus they all have the same value a(2) = a(3) = a(6) = a(30) = 2 in this sequence.
Number 4 (2^2), 9 (3^2) and 36 (2^2 * 3^2) all have just one distinct exponent, 2, in the multisets of exponents that occur in their prime factorization, thus they all have the same value a(4) = a(9) = a(36) = 3 in this sequence.
Numbers 12 = 2^2 * 3^1, 18 = 2^1 * 3^2, 60 = 2^2 * 3^1 * 5^1 and 300 = 2^2 * 3^1 * 5^2 all have both 1 and 2 and none other values occurring in the multisets of exponents in their prime factorization, thus they all have the value of a(12) = 5 that was allotted to 12 by the restricted growth sequence transform, as 12 is the smallest number with prime signature (1, 2).
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PROG
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(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; };
A007947(n) = factorback(factorint(n)[, 1]);
A181819(n) = factorback(apply(e->prime(e), (factor(n)[, 2])));
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CROSSREFS
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Cf. A005117 (gives indices of terms <= 2), A062503 (after its initial 1, gives indices of 3's).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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