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%I #53 Feb 09 2024 08:41:40
%S 1,0,0,0,0,0,0,0,1,0,0,1,0,0,1,1,0,0,0,1,1,0,0,0,1,0,0,1,0,0,0,0,1,0,
%T 1,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,1,1,0,0,1,0,1,0,0,0,0,0,0,0,1,0,0,1,
%U 1,0,0,0,0,0,0,1,1,0,0,0,1,0,0,0,1,0,1,0,0,0,1,1,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,1,1,0,0,1,0,1
%N a(n) = 1 if n' / gcd(n,n') is even, otherwise 0, where n' stands for the arithmetic derivative of n, A003415.
%C a(n) = 1 if A083345(n) = numerator of Sum(e/p: n=Product(p^e)) is even, and 0 if it is odd.
%C Question: Is the asymptotic mean of this sequence 1/3? See also A369004.
%C Answer to the above question is yes, as 1/4 + 1/16 + 1/64 + 1/256 + 1/1024 + 1/4096 + ... = 1/3. See the new recursive formula, whose first term contributes 1/4, and the second term 1/16 to the total asymptotic mean, with the rest obtained by recursion. For a proof, consider A001787(n) = A003415(2^n) = n*2^(n-1). We have A007814(A001787(n)) > n iff n is a multiple of 4. - _Antti Karttunen_, Jan 29 2024
%C Also a(n) = 1 iff A276085(n) is a multiple of 4. See comment in A327860, which applies also to A342002 [= A083345(A276086(n))]. Therefore, A121262(n) = A059841(A342002(n)) = A059841(A083345(A276086(n))) = a(A276086(n)). Assuming that the new formula a(n) = A121262(A276085(n)) holds, then substituing A276086(n) for n in it gives us back equation a(A276086(n)) = A121262(n) that was proved above. - _Antti Karttunen_, Feb 08 2024
%H Antti Karttunen, <a href="/A369001/b369001.txt">Table of n, a(n) for n = 1..65537</a>
%H <a href="/index/Ch#char_fns">Index entries for characteristic functions</a>
%F a(n) = A059841(A083345(n)).
%F a(n) >= A369004(n).
%F A353557(n) <= a(n) <= A368997(n) <= A358680(n).
%F From _Antti Karttunen_, Jan 29 2024 and Feb 08 2024: (Start)
%F For n > 1, a(n) = [A007814(A003415(n)) > A007814(n)], where [ ] is the Iverson bracket.
%F a(n) = A353557(n) + [A007814(n)==2]*A066829(n) + [A007814(n)>=4]*a(n/16).
%F a(n) = a(16*n).
%F a(n) <= A035263(n).
%F a(n) = A121262(A276085(n)). [See comments]
%F (End)
%o (PARI)
%o A083345(n) = { my(f=factor(n)); numerator(vecsum(vector(#f~, i, f[i, 2]/f[i, 1]))); };
%o A369001(n) = !(A083345(n)%2);
%o (PARI)
%o A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
%o A369001(n) = ((1==n)||(valuation(A003415(n),2) > valuation(n,2)));
%o (PARI) A369001(n) = (((n%2)&&(!(bigomega(n)%2))) || (2==valuation(n,2) && (bigomega(n)%2)) || (valuation(n,2)>=4 && A369001(n/16)));
%Y Characteristic function of A369002, whose complement A369003 gives the positions of 0's.
%Y Cf. A001787, A003415, A007814, A035263, A059841, A066829, A083345, A085731, A121262 [= a(A276086(n))], A276085, A276086, A327860, A342002, A353557, A359820, A369004, A369643.
%Y Cf. also A358680, A368997.
%K nonn
%O 1
%A _Antti Karttunen_, Jan 14 2024
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