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%I #16 May 13 2018 20:45:20
%S 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,
%T 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,
%U 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
%N Partial sums of f(n) = (-1)^(1-A304109(n)).
%C 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.
%C The first negative term occurs as a(153) = -1. See also comments at A304010.
%H Antti Karttunen, <a href="/A304111/b304111.txt">Table of n, a(n) for n = 0..65537</a>
%H <a href="/index/Ge#GF2X">Index entries for sequences related to polynomials in ring GF(2)[X]</a>
%F a(0) = 0, and for n > 0, a(n) = a(n-1) + (-1)^(1-A304109(n)).
%F For n >= 1, a(n) = (2*A304110(n)) - n.
%o (PARI)
%o up_to = 128;
%o A304109(n) = { my(fm=factor(Pol(binary(n))*Mod(1, 2))); for(k=1, #fm~, if(fm[k, 2] > 1, return(0))); (1); };
%o prepare_v304110(up_to) = { my(v=vector(up_to), c=0); for(n=1, up_to, c += A304109(n); v[n] = c); (v); };
%o v304110 = prepare_v304110(up_to);
%o A304110(n) = v304110[n];
%o A304111(n) = ((2*A304110(n)) - n);
%o \\ Or just as:
%o c=0; for(n=0, up_to, if(n>0, c+=((-1)^(1-A304109(n)))); print1(c, ", "));
%Y Cf. A304107, A304108, A304109, A304110.
%K sign
%O 0,3
%A _Antti Karttunen_, May 13 2018
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