A197774 Suppose n has prime factorization n = p1^a1 * p2^a2 * ... * pk^ak. Then a(n) = (-1)^(n1 + n2 + ... + nk) if all the ai are ni^2 and a(n) = 0 otherwise.

1, -1, -1, 0, -1, 1, -1, 0, 0, 1, -1, 0, -1, 1, 1, 1, -1, 0, -1, 0, 1, 1, -1, 0, 0, 1, 0, 0, -1, -1, -1, 0, 1, 1, 1, 0, -1, 1, 1, 0, -1, -1, -1, 0, 0, 1, -1, -1, 0, 0, 1, 0, -1, 0, 1, 0, 1, 1, -1, 0, -1, 1, 0, 0, 1, -1, -1, 0, 1, -1, -1, 0, -1, 1, 0, 0, 1, -1, -1, -1, 1, 1, -1, 0, 1, 1, 1, 0, -1, 0, 1, 0, 1, 1, 1, 0, -1, 0, 0, 0, -1, -1, -1, 0, -1

Offset

1

Comments

Differs from A219009 at n=32, 96, 160, 224, 243, 256, 352, ... - R. J. Mathar, May 28 2016

Links

Antti Karttunen, Table of n, a(n) for n = 1..10000

Mathematics StackExchange, Two Dirichlet's series related to the Divisor Summatory Function and to the Riemann's zeta-function.

Index entries for sequences computed from exponents in factorization of n.

Formula

Multiplicative with a(p^e) = (-1)^sqrt(e) if e is a square, 0 otherwise. - Franklin T. Adams-Watters, Oct 18 2011

From Amiram Eldar, Nov 09 2023:

a(n) = (-1)^A001222(n) if n is in A197680 and 0 otherwise.

Limit_{m->oo} (1/m) * Sum_{k=1..m} abs(a(k)) = 0.64111516... (A357016). (End)

Example

From Michael De Vlieger, Jul 24 2017: (Start)
a(5) = -1 since 5^1 has an exponent 1 that is a perfect square, thus (-1)^sqrt(1) = -1.
a(6) = 1 since 6 = 2^1 * 3^1; both exponents are perfect squares thus (-1)^sqrt(1) * (-1)^sqrt(1) = -1 * -1 = 1.
a(12) = 0 since 12 = 2^2 * 3^1. One exponent (1) is a perfect square but the other (2) is not, thus 0 * (-1)^sqrt(1) = 0.
(End)

Maple

A197774 := proc(n)
    local a, pf, e ;
    a := 1 ;
    for pf in ifactors(n)[2] do
        e := pf[2] ;
        if issqr(e) then
            a := a*(-1)^sqrt(e) ;
        else
            a := 0 ;
        end if;
    end do;
    a;
end proc: # R. J. Mathar, May 28 2016

Mathematica

Table[If[n == 1, 1, Apply[Times, FactorInteger[n] /. {p_, e_} /; p > 0 :> If[IntegerQ@ Sqrt@ e, (-1)^Sqrt@ e, 0]]], {n, 105}] (* Michael De Vlieger, Jul 24 2017 *)

Prog

(PARI) A197774(n) = { my(f=factor(n)[, 2]); prod(i=1, #f, if(issquare(f[i]), (-1)^sqrtint(f[i]), 0)); }; \\ Antti Karttunen, Jul 24 2017

Crossrefs

Cf. A001222, A197680, A219009, A357016.

Sequence in context: A133639 A323154 A060038 * A219009 A353519 A267708

Adjacent sequences: A197771 A197772 A197773 * A197775 A197776 A197777

Keyword

sign,easy,mult

Author

A. Neves, Oct 18 2011

Extensions

More terms from Antti Karttunen, Jul 24 2017

Status

approved