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A304111
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Partial sums of f(n) = (-1)^(1-A304109(n)).
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4
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0, 1, 2, 3, 2, 1, 2, 3, 2, 3, 2, 3, 2, 3, 4, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 3, 4, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 4, 3, 2, 3, 2, 3, 2, 1, 2, 3, 2, 3, 4, 3, 2, 3, 2, 3, 2, 1, 2, 3, 2, 3, 4, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 3, 4, 3, 2, 3, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 3, 2, 3, 2, 3, 4, 3, 2, 3, 4, 3, 2, 3, 2, 3, 2, 1, 2, 1, 0, 1, 2, 3, 2, 3, 2, 3, 2, 3, 4, 3, 2
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OFFSET
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0,3
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COMMENTS
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Start from the initial value a(0) = 0, after which, for n > 0, each successive term a(n) is either one more or one less than the previous term a(n-1), depending on whether the binary expansion of n encodes a squarefree (0,1)-polynomial when the factorization is done in polynomial ring GF(2)[X]], or whether it encodes a polynomial where at least one of its irreducible divisors occurs more than once.
The first negative term occurs as a(153) = -1. See also comments at A304010.
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LINKS
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FORMULA
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a(0) = 0, and for n > 0, a(n) = a(n-1) + (-1)^(1-A304109(n)).
For n >= 1, a(n) = (2*A304110(n)) - n.
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PROG
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(PARI)
up_to = 128;
A304109(n) = { my(fm=factor(Pol(binary(n))*Mod(1, 2))); for(k=1, #fm~, if(fm[k, 2] > 1, return(0))); (1); };
prepare_v304110(up_to) = { my(v=vector(up_to), c=0); for(n=1, up_to, c += A304109(n); v[n] = c); (v); };
v304110 = prepare_v304110(up_to);
\\ Or just as:
c=0; for(n=0, up_to, if(n>0, c+=((-1)^(1-A304109(n)))); print1(c, ", "));
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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