|
%I #38 Jan 09 2023 12:08:24
%S 1,2,3,1,5,6,7,2,3,10,11,3,13,14,15,1,17,6,19,5,21,22,23,6,5,26,1,7,
%T 29,30,31,2,33,34,35,3,37,38,39,10,41,42,43,11,15,46,47,3,7,10,51,13,
%U 53,2,55,14,57,58,59,15,61,62,21,1,65,66,67,17,69,70,71,6,73,74,15,19,77,78
%N Denominator of r(n) = Sum(e/p: n=Product(p^e)).
%C Multiplicative with a(p^e) = 1 iff p|e, p otherwise. For f(n) = A083345(n)/A083346(n), f(p^i*q^j*...) = f(p^i)+f(q^j)+ ... The denominator of each term is 1 or the prime, thus the denominator of the sum is the product of the denominators of the components. - _Christian G. Bower_, May 16 2005
%C n divided by the greatest common divisor of n and its arithmetic derivative, i.e., a(n) = n/gcd(n,n') = A000027(n)/A085731(n). - _Giorgio Balzarotti_, Apr 14 2011
%H Antti Karttunen, <a href="/A083346/b083346.txt">Table of n, a(n) for n = 1..65537</a>
%F Sum_{k=1..n} a(k) ~ c * n^2, where c = A065463 * Product_{p prime} (p^(2*p)*(p^2+p-1)-p^3)/((p^2+p-1)*(p^(2*p)-1)) = 0.3374565531... . - _Amiram Eldar_, Nov 18 2022
%e n=12 = 2*2*3 = 2^2 * 3^1 -> r(12) = 2/2 + 1/3 = (6+2)/6, therefore a(12)=3, A083345(12)=4;
%e n=18 = 2*3*3 = 2^1 * 3^2 -> r(18) = 1/2 + 2/3 = (3+4)/6, therefore a(18)=6, A083345(18)=7.
%t a[n_] := Product[Module[{p, e}, {p, e} = pe; If[Divisible[e, p], 1, p]], {pe, FactorInteger[n]}];
%t Array[a, 100] (* _Jean-François Alcover_, Oct 06 2021 *)
%o (PARI) A083346(n) = { my(f=factor(n)); denominator(vecsum(vector(#f~,i,f[i,2]/f[i,1]))); }; \\ _Antti Karttunen_, Mar 01 2018
%Y Cf. A083345 (numerator).
%Y Cf. A035263 (parity of terms), A003159 (positions of odd terms), A036554 (of even terms).
%Y Cf. A065463, A072873, A083347, A083348, A359588 (Dirichlet inverse).
%K nonn,mult
%O 1,2
%A _Reinhard Zumkeller_, Apr 25 2003
%E Incorrect formula removed by _Antti Karttunen_, Jan 09 2023
|
