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%I #17 Jun 12 2022 15:15:48
%S 1,2,3,4,5,3,7,8,9,5,11,6,13,7,5,16,17,18,19,5,21,11,23,12,25,13,27,
%T 14,29,10,31,32,11,17,35,36,37,19,39,10,41,42,43,22,15,23,47,24,49,50,
%U 17,13,53,27,55,28,57,29,59,20,61,31,63,64,65,22
%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) = denominator of s. a(1)=1 by convention.
%H Seiichi Manyama, <a href="/A322035/b322035.txt">Table of n, a(n) for n = 1..16384</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) = 6.
%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
%Y Cf. A006022, A027746, A322034, A322036.
%Y A017665/A017666 = sum of reciprocals of all divisors of n.
%K nonn,frac
%O 1,2
%A _N. J. A. Sloane_ and _David James Sycamore_, Nov 28 2018