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Let p1 <= p2 <= ... <= pk be the prime factors of n, with repetition; let s = 1/p1 + 1/(p1*p2) + 1/(p1*p2*p3) + ... + 1/(p1*p2*...*pk); a(n) = numerator of s. a(1)=0 by convention.
5

%I #33 Jun 07 2019 16:30:31

%S 0,1,1,3,1,2,1,7,4,3,1,5,1,4,2,15,1,13,1,4,8,6,1,11,6,7,13,11,1,7,1,

%T 31,4,9,8,31,1,10,14,9,1,29,1,17,7,12,1,23,8,31,6,10,1,20,12,25,20,15,

%U 1,17,1,16,29,63,14,15,1,13,8,43,1,67

%N Let p1 <= p2 <= ... <= pk be the prime factors of n, with repetition; let s = 1/p1 + 1/(p1*p2) + 1/(p1*p2*p3) + ... + 1/(p1*p2*...*pk); a(n) = numerator of s. a(1)=0 by convention.

%C Note that s < 1 for all n (compare A322036). This follows easily by induction, since when we increase n by multiplying it by a new (not-smaller) prime, we increase s by less than 1-s.

%H Antti Karttunen, <a href="/A322034/b322034.txt">Table of n, a(n) for n = 1..16384</a>

%H Antti Karttunen, <a href="/A322034/a322034.txt">Data supplement: n, a(n) computed for n = 1..65537</a>

%e If n=12 we get the prime factors 2,2,3, and s = 1/2 + 1/4 + 1/12 = 5/6. So a(12) = 5.

%e The fractions s for n >= 2 are 1/2, 1/3, 3/4, 1/5, 2/3, 1/7, 7/8, 4/9, 3/5, 1/11, 5/6, 1/13, 4/7, 2/5, 15/16, 1/17, 13/18, 1/19, 4/5, 8/21, ...

%p # This generates the terms starting at n=2:

%p P:=proc(n) local FM: FM:=ifactors(n)[2]: seq(seq(FM[j][1], k=1..FM[j][2]), j=1..nops(FM)) end: # A027746

%p f0:=[]; f1:=[]; f2:=[];

%p for n from 2 to 120 do

%p a:=0; b:=1; t1:=[P(n)];

%p for i from 1 to nops(t1) do b:=b/t1[i]; a:=a+b; od;

%p f0:=[op(f0),a]; f1:=[op(f1), numer(a)]; f2:=[op(f2),denom(a)]; od:

%p f0; # s

%p f1; # A322034

%p f2; # A322035

%p f2-f1; # A322036

%o (PARI) A322034(n) = if(1==n,0,my(f=factor(n),pm=1,s=0); for(i=1,#f~,while(f[i,2],pm *= f[i,1]; f[i,2]--; s += 1/pm)); numerator(s)); \\ _Antti Karttunen_, Feb 28 2019

%Y Cf. A006022, A027746, A322035, A322036.

%Y A017665/A017666 = sum of reciprocals of all divisors of n.

%K nonn,frac

%O 1,4

%A _N. J. A. Sloane_ and _David James Sycamore_, Nov 28 2018