A260685 Sequence is defined by the condition that Sum_{d|n} a(d)^(n/d) = 1 if n=1, = 0 if n>1.

1, -1, -1, -2, -1, -1, -1, -6, 0, -1, -1, 4, -1, -1, 1, -54, -1, 0, -1, 28, 1, -1, -1, 132, 0, -1, 0, 124, -1, -1, -1, -4470, 1, -1, 1, 444, -1, -1, 1, 5964, -1, -1, -1, 2044, 0, -1, -1, 89028, 0, 0, 1, 8188, -1, 0, 1, 248172, 1, -1, -1, 9784, -1, -1, 0, -30229110

Offset: 1

Author

Gevorg Hmayakyan, Nov 15 2015

Keywords

• sign

Comments

It is easy to prove that a(1)=1, a(p)=-1, a(p^n)=0 if p>2, a(p1*p2*..*pn)=(-1)^n, a(2*p1*...*pn)=-1.

It appears that abs(a(n)) > 1 for multiples of 4. - Michel Marcus, Nov 19 2015

If p1,...,pn are odd it appears that a(p1^k1*p2^k2*...*pn^kn)=0 if one of k1,...,kn > 1. Similarly, it appears that a(2*p1^k1*p2^k2*...*pn^kn)=0 if one of k1,...,kn > 1.

For odd n a(n) is equal to the Möbius function: A008683(n).

For n == 2 mod 4, it seems that a(n) = -|Möbius(n/2)|. For n == 0 mod 4, see A264609.- N. J. A. Sloane, Nov 24 2015

Examples

			For a prime p, a(p)^1 + a(1)^p = 0 => a(p) = -1.
For n=6, a(1)^6 + a(2)^3 + a(3)^2 + a(6)^1 = 0, so 1 - 1 + 1 + a(6) = 0, so 1 + a(6) = 0, so a(6) = -1.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..8447

Crossrefs

Cf. A008683, A264609 (a(4n)), A264610 (a(2^n)).

Programs

Maple

a:= proc(n) option remember;
    -add(procname(n/d)^d, d = numtheory:-divisors(n) minus {1});
end proc:
a(1):= 1:
map(a, [$1..100]); # Robert Israel, Nov 19 2015

Mathematica

a[1] = 1; a[n_] := a[n] = -DivisorSum[n, If[# == 1, 0, a[n/#]^#] &]; Array[a, 70] (* Jean-François Alcover, Dec 02 2015, adapted from PARI *)

PARI

a(n) = if (n==1, 1, - sumdiv(n, d, if (d==1, 0, a(n/d)^d))); \\ Michel Marcus, Nov 16 2015