A304111 Partial sums of f(n) = (-1)^(1-A304109(n)).

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

Offset: 0

Author

Antti Karttunen, May 13 2018

Keywords

• sign

Comments

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.

Links

• Antti Karttunen, Table of n, a(n) for n = 0..65537
• Index entries for sequences related to polynomials in ring GF(2)[X]

Crossrefs

Cf. A304107, A304108, A304109, A304110.

Programs

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);
A304110(n) = v304110[n];
A304111(n) = ((2*A304110(n)) - n);
\\ Or just as:
c=0; for(n=0, up_to, if(n>0, c+=((-1)^(1-A304109(n)))); print1(c, ", "));

Formula

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.