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Nov5-11, 06:58 AM
P: 15,201
I'll repeat, what's wrong with "divides"? The sequence {a divides n} for a given a and n comprising the positive integers is the opposite of what you want. Take the logical negation and voila! there is your set of sequences.

By the way, you have a missing sequence, the trivial sequence, in your set of sequences. It is the sequence 0,0,0,... So let's start with this in terms of "divides". 1 divides every positive integer, so the sequence given by {dn;1} where dn;1 is 1 divides n is the sequence of all ones (or all true). The logical negation of this sequence yields the trivial series 0,0,0,... .

Now look at using 2 as the divisor. 2 does not divide 1, it divides 2, it does not divide 3, and so on. Thus {dn;2} = {2 divides n} yields 0,1,0,1,... . The logical negation of this sequence yields 1,0,1,0,... : your first example.

Now look at using 3 the divisor. 3 divides n results in the sequence 0,0,1,0,0,1,... . The logical negation yields 1,1,0,1,1,0 : your second example. Doing the same with 4 yields 1,1,1,0,1,1,1,0,...

Divides (or does not divide) is exactly what you want here.