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A226774
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Integers a(n) = Sum_{i=1..q} 1/d(i) where d(i) are the divisors of A225110(n) for some q.
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1
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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, 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, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
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OFFSET
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1,2
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COMMENTS
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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, ...
By convention, a(1)=1. For a majority of n, a(n) = 2.
a(n) = 3 for n = 11, 16, 52, 145, 634, ... where A225110(n) = 120, 180, 672, 1890, 8460, ...
a(n) = 4 for n = 2284, 2476, 6871, ... where A225110(n) = 30240, 32760, 90720, ...
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LINKS
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EXAMPLE
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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.
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MAPLE
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with(numtheory): for n from 1 to 2000 do:x:=divisors(n):n1:=nops(x):s:=0:ii:=0:for q from 1 to
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:
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PROG
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(PARI)
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); };
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)));
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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