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%I #14 Dec 17 2017 03:08:06
%S 1,2,2,2,2,2,2,2,2,2,3,2,2,2,2,3,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,
%T 2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,
%U 2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2
%N Integers a(n) = Sum_{i=1..q} 1/d(i) where d(i) are the divisors of A225110(n) for some q.
%C The corresponding q are 1, 4, 4, 6, 4, 4, 4, 4, 4, 4, 16, 4, 4, 4, 4, 15, 4, 6, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 10, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 24, 4, 4, 4, 4, 4, 4, ...
%C By convention, a(1)=1. For a majority of n, a(n) = 2.
%C a(n) = 3 for n = 11, 16, 52, 145, 634, ... where A225110(n) = 120, 180, 672, 1890, 8460, ...
%C a(n) = 4 for n = 2284, 2476, 6871, ... where A225110(n) = 30240, 32760, 90720, ...
%H Antti Karttunen, <a href="/A226774/b226774.txt">Table of n, a(n) for n = 1..16384</a>
%e a(16) = 3 because the divisors of A225110(16) = 180 are 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180 and 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/9 + 1/10 + 1/12 + 1/15 + 1/18 + 1/20 + 1/30 + 1/36 + 1/45 = 3.
%p with(numtheory): for n from 1 to 2000 do:x:=divisors(n):n1:=nops(x):s:=0:ii:=0:for q from 1 to
%p n1 while(ii=0) do:s:=s+1/x[q]:if s=floor(s) and q>1 then ii:=1: printf(`%d, `, s):else fi:od:od:
%o (PARI)
%o either_A226774_or_0(n) = { if(1==n,return(1)); my(divs=divisors(n),s=0); for(i=1,#divs,s += (1/divs[i]); if((1==denominator(s))&&(i>1),return(s))); return(0); };
%o up_to = 16384; i=0; n=0; while(i<up_to, n++; s = either_A226774_or_0(n); if(s, i++; write("b226774.txt", i, " ", s)));
%o \\ _Antti Karttunen_, Dec 16 2017
%Y Cf. A225110.
%K nonn
%O 1,2
%A _Michel Lagneau_, Jun 17 2013
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