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If n = Product (p_j^k_j) then a(n) = denominator of Product (1 + 1/p_j^k_j).
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%I #18 Nov 13 2021 10:21:48

%S 1,2,3,4,5,1,7,8,9,5,11,3,13,7,5,16,17,3,19,2,21,11,23,2,25,13,27,7,

%T 29,5,31,32,11,17,35,18,37,19,39,20,41,7,43,11,3,23,47,12,49,25,17,26,

%U 53,9,55,7,57,29,59,1,61,31,63,64,65,11,67,34,23,35

%N If n = Product (p_j^k_j) then a(n) = denominator of Product (1 + 1/p_j^k_j).

%C Denominator of sum of reciprocals of unitary divisors of n.

%H Antti Karttunen, <a href="/A332883/b332883.txt">Table of n, a(n) for n = 1..20000</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/UnitaryDivisor.html">Unitary Divisor</a>

%F a(n) = denominator of Sum_{d|n, gcd(d, n/d) = 1} 1/d.

%F a(n) = denominator of usigma(n)/n.

%F a(p) = p, where p is prime.

%F a(n) = n / A323166(n). - _Antti Karttunen_, Nov 13 2021

%e 1, 3/2, 4/3, 5/4, 6/5, 2, 8/7, 9/8, 10/9, 9/5, 12/11, 5/3, 14/13, 12/7, 8/5, 17/16, ...

%p a:= n-> denom(mul(1+i[1]^i[2], i=ifactors(n)[2])/n):

%p seq(a(n), n=1..80); # _Alois P. Heinz_, Feb 28 2020

%t Table[If[n == 1, 1, Times @@ (1 + 1/#[[1]]^#[[2]] & /@ FactorInteger[n])], {n, 1, 70}] // Denominator

%t Table[Sum[If[GCD[d, n/d] == 1, 1/d, 0], {d, Divisors[n]}], {n, 1, 70}] // Denominator

%o (PARI) a(n) = denominator(sumdiv(n, d, if (gcd(d, n/d)==1, 1/d))); \\ _Michel Marcus_, Feb 28 2020

%Y Cf. A007947, A017666, A034448, A077610, A319677, A323166, A327158 (positions of 1's), A332881, A332882 (numerators).

%K nonn,frac

%O 1,2

%A _Ilya Gutkovskiy_, Feb 28 2020