A262804 a(n) = 2*b(n), where b(n) is defined by the condition that Product_{d|n} (b(d) + 1) = 1, n > 1 and b(1) = 1.

2, -1, -1, 0, -1, 2, -1, 0, 0, 2, -1, 0, -1, 2, 2, 0, -1, 0, -1, 0, 2, 2, -1, 0, 0, 2, 0, 0, -1, -1, -1, 0, 2, 2, 2, 0, -1, 2, 2, 0, -1, -1, -1, 0, 0, 2, -1, 0, 0, 0, 2, 0, -1, 0, 2, 0, 2, 2, -1, 0, -1, 2, 0, 0, 2, -1, -1, 0, 2, -1, -1, 0, -1, 2, 0, 0, 2, -1

Offset

1,1

Comments

We can see from the formula that:

Product_{d|n} (moebius(d)*(moebius(d) + 3) + 4) = 4^d(n), for n > 1, where moebius(n) and d(n) are A008683 and A000005 respectively.

It also can be proved that

Product_{d|n} ((moebius(d) - 3/2 + I*sqrt(7)/2)) = Product_{d|n} ((moebius(d) - 3/2 - I*sqrt(7)/2)) is integer iff n belongs to A048109, where I = sqrt(-1).

Links

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

Index entries for sequences computed from exponents in factorization of n

Formula

a(p) = -1 if p is prime.

a(n) = 2*moebius(n) if moebius(n) = 1;

a(n) = moebius(n) if moebius(n) != 1.

And in general:

a(n) = moebius(n)*(moebius(n) + 3)/2.

Example

If p is prime, (b(p)+1)(b(1)+1) = 1 => b(p)+1 = 1/2 => b(p) = -1/2 => a(p) = 2*b(p) = -1.
For n=6, (1+b(1))(1+b(2))(1+b(3))(1+b(6)) = 1 => 2*(1/2)*(1/2)*(1+b(6)) = 1 => b(6) = 1 => a(6) = 2*b(6) = 2. - Example corrected by Antti Karttunen, Jul 25 2017

Maple

b:= proc(n) option remember; if n>1 then (1)/(mul((1+b(d)), d = numtheory:-divisors(n) minus {n}))-1 else 1; end if;  end proc:  b(1):= 1:a(i):=2*b(i):  L:=seq(a(i), i=1..200);

Mathematica

Table[MoebiusMu[n] (MoebiusMu[n] + 3) / 2, {n, 80}] (* Vincenzo Librandi, Jan 19 2016 *)

Prog

(PARI) a(n) = moebius(n)*(moebius(n)+3)/2; \\ Michel Marcus, Jan 19 2016
(Magma) [MoebiusMu(n)*(MoebiusMu(n)+3)/2: n in [1..100]]; // Vincenzo Librandi, Jan 19 2016

Crossrefs

Cf. A000005, A008683, A048109.

Sequence in context: A346392 A169590 A300185 * A048825 A339884 A116375

Adjacent sequences: A262801 A262802 A262803 * A262805 A262806 A262807

Keyword

sign

Author

Gevorg Hmayakyan, Jan 17 2016

Status

approved