|
%I #16 Sep 03 2021 14:10:33
%S 0,1,-1,0,1,-1,-1,1,0,0,1,-1,-1,-1,1,0,1,1,-1,1,0,0,1,-1,0,-1,-1,-1,
%T -1,0,1,1,1,0,-1,0,-1,-1,0,0,1,1,-1,1,1,0,1,-1,0,1,1,-1,-1,-1,0,-1,0,
%U -1,1,1,-1,0,-1,0,-1,0,1,1,1,-1,-1,1,1,-1,-1,-1,1,1,-1,1,0,0,1,0,0,-1,0,0,-1,0,0,1,1,0,-1,-1,1,1,1,0,-1,0,1,-1,0
%N a(n) is -1, 0, or +1 such that a(n) == A156552(n) (mod 3).
%H Antti Karttunen, <a href="/A332814/b332814.txt">Table of n, a(n) for n = 1..65537</a>
%H <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>
%F a(n) = A102283(A156552(n)).
%F If A329903(n) = 2, then a(n) = -1, otherwise a(n) = A329903(n).
%F a(n) = A332823(A332461(n)) = A332823(A332462(n)).
%F a(2^n) = A000035(n), for all n >= 0.
%F a(n^2) = 0, for all n >= 1.
%F a(A000040(n)) = (-1)^(n-1).
%F a(A003961(n)) = -a(n).
%F a(2*n) = ((2-a(n)) mod 3) - 1. - _Peter Munn_, Aug 28 2021
%o (PARI)
%o A156552(n) = {my(f = factor(n), p2 = 1, res = 0); for(i = 1, #f~, p = 1 << (primepi(f[i, 1]) - 1); res += (p * p2 * (2^(f[i, 2]) - 1)); p2 <<= f[i, 2]); res}; \\ From A156552
%o A332814(n) = { my(u=A156552(n)%3); if(2==u,-1,u); };
%Y Cf. A000040, A003961, A102283, A156552, A329903, A332461, A332462, A332823.
%Y Cf. A329609, A329604, A332812 for positions of 0's, +1's and -1's in this sequence.
%K sign
%O 1
%A _Antti Karttunen_, Mar 01 2020
|
