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A369001
a(n) = 1 if n' / gcd(n,n') is even, otherwise 0, where n' stands for the arithmetic derivative of n, A003415.
24
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, 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, 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
OFFSET
1
COMMENTS
a(n) = 1 if A083345(n) = numerator of Sum(e/p: n=Product(p^e)) is even, and 0 if it is odd.
Question: Is the asymptotic mean of this sequence 1/3? See also A369004.
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
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
FORMULA
a(n) = A059841(A083345(n)).
a(n) >= A369004(n).
A353557(n) <= a(n) <= A368997(n) <= A358680(n).
From Antti Karttunen, Jan 29 2024 and Feb 08 2024: (Start)
For n > 1, a(n) = [A007814(A003415(n)) > A007814(n)], where [ ] is the Iverson bracket.
a(n) = A353557(n) + [A007814(n)==2]*A066829(n) + [A007814(n)>=4]*a(n/16).
a(n) = a(16*n).
a(n) <= A035263(n).
a(n) = A121262(A276085(n)). [See comments]
(End)
a(n) = A035263(n) - A373141(n) = A369004(n) + A373264(n) = A373137(n) + A373266(n). - Antti Karttunen, Jun 01 2024
PROG
(PARI)
A083345(n) = { my(f=factor(n)); numerator(vecsum(vector(#f~, i, f[i, 2]/f[i, 1]))); };
A369001(n) = !(A083345(n)%2);
(PARI)
A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
A369001(n) = ((1==n)||(valuation(A003415(n), 2) > valuation(n, 2)));
(PARI) A369001(n) = (((n%2)&&(!(bigomega(n)%2))) || (2==valuation(n, 2) && (bigomega(n)%2)) || (valuation(n, 2)>=4 && A369001(n/16)));
CROSSREFS
Characteristic function of A369002, whose complement A369003 gives the positions of 0's.
Cf. A369974 (Dirichlet inverse), A369975 (and its parity), A369980 (run lengths).
Sequence in context: A324917 A369974 A369975 * A361024 A354037 A185118
KEYWORD
nonn
AUTHOR
Antti Karttunen, Jan 14 2024
STATUS
approved