|
%I #53 Dec 02 2015 06:37:34
%S 1,-1,-1,-2,-1,-1,-1,-6,0,-1,-1,4,-1,-1,1,-54,-1,0,-1,28,1,-1,-1,132,
%T 0,-1,0,124,-1,-1,-1,-4470,1,-1,1,444,-1,-1,1,5964,-1,-1,-1,2044,0,-1,
%U -1,89028,0,0,1,8188,-1,0,1,248172,1,-1,-1,9784,-1,-1,0,-30229110
%N Sequence is defined by the condition that Sum_{d|n} a(d)^(n/d) = 1 if n=1, = 0 if n>1.
%C 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.
%C It appears that abs(a(n)) > 1 for multiples of 4. - _Michel Marcus_, Nov 19 2015
%C 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.
%C For odd n a(n) is equal to the Möbius function: A008683(n).
%C 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
%H Robert Israel, <a href="/A260685/b260685.txt">Table of n, a(n) for n = 1..8447</a>
%e For a prime p, a(p)^1 + a(1)^p = 0 => a(p) = -1.
%e 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.
%p a:= proc(n) option remember;
%p -add(procname(n/d)^d, d = numtheory:-divisors(n) minus {1});
%p end proc:
%p a(1):= 1:
%p map(a, [$1..100]); # _Robert Israel_, Nov 19 2015
%t 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 *)
%o (PARI) a(n) = if (n==1, 1, - sumdiv(n, d, if (d==1, 0, a(n/d)^d))); \\ _Michel Marcus_, Nov 16 2015
%Y Cf. A008683, A264609 (a(4n)), A264610 (a(2^n)).
%K sign
%O 1,4
%A _Gevorg Hmayakyan_, Nov 15 2015
|
