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If n = Product (p_j^k_j) then a(n) = numerator of Product (1 + 1/p_j^k_j).
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%I #23 Nov 21 2022 09:39:12

%S 1,3,4,5,6,2,8,9,10,9,12,5,14,12,8,17,18,5,20,3,32,18,24,3,26,21,28,

%T 10,30,12,32,33,16,27,48,25,38,30,56,27,42,16,44,15,4,36,48,17,50,39,

%U 24,35,54,14,72,9,80,45,60,2,62,48,80,65,84,24,68,45,32,72

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

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

%H Antti Karttunen, <a href="/A332882/b332882.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) = numerator of Sum_{d|n, gcd(d, n/d) = 1} 1/d.

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

%F a(p) = p + 1, where p is prime.

%F a(n) = A034448(n) / A323166(n). - _Antti Karttunen_, Nov 13 2021

%F Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k)/A332883(k) = zeta(2)/zeta(3) = 1.368432... (A306633). - _Amiram Eldar_, Nov 21 2022

%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-> numer(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}] // Numerator

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

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

%Y Cf. A017665, A028235, A028236, A034448, A077610, A306633, A319676, A323166, A332880, A332883 (denominators).

%Y Cf. also A348734, A348735.

%K nonn,frac

%O 1,2

%A _Ilya Gutkovskiy_, Feb 28 2020