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Difference between the number of nonnegative evil and odious multiples of n less than 2^n.
1

%I #15 Apr 23 2022 09:42:42

%S -1,-1,3,-1,5,9,-7,-1,27,25,11,162,13,49,521,-1,697,2196,19,625,8435,

%T 121,-23,59049,3120,169,177147,2401,29,982635,-237367,-1,3958307,

%U 781745,121691,28697814,37,361,89405461,1953125

%N Difference between the number of nonnegative evil and odious multiples of n less than 2^n.

%C It appears that for n = 2^i*p, i>0, p prime, a(n) = p^(2^i) or a multiple of it.

%C For which n is a(n) negative?

%C For prime n and related conjectures, see A133954.

%o (PARI) a(n)=sum(i=0,floor(2^n/n),subst(Pol(binary(i*n)),x,1)%2==0)-sum(i=0,floor(2^n/n),subst(Pol(binary(i*n)),x,1)%2)

%Y Cf. A000069, A001969, A133954.

%K sign,hard,more

%O 1,3

%A _Ralf Stephan_, Aug 31 2013

%E a(31)-a(40) from _Amiram Eldar_, Apr 23 2022