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%I #19 Sep 17 2017 18:40:28
%S 1,1,2,3,6,1,2,3,6,1,2,4,7,14,28,1,2,3,6,1,2,3,6,1,2,3,6,1,2,3,6,1,2,
%T 3,6,1,2,3,6,1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120,1,2,3,6,1,2,3,
%U 6,1,2,3,6,1,2,3,6,1,2,3,4,5,6,9,10,12,15,18
%N Triangle read by rows: row n gives the first q divisors d(1), d(2), ..., d(q) of A225110(n) such that Sum_{i = 1..q} 1/d(i) is an integer.
%C Rows 2, 3, 5, 6, 7, ... with the divisors {1, 2, 3, 6} are identical;
%C rows 4, 18, 62, 67, ... with the divisors {1, 2, 4, 7, 14, 28} are identical;
%C ...
%C The primitive rows are rows 1, 2, 4, 11, 16, 39, 52, 145, ... corresponding to n = 1, 6, 28, 120, 180, 496, 672, 1890, ... (see A226853).
%C The irregular triangle of divisors is:
%C [1]
%C [1, 2, 3, 6]
%C [1, 2, 3, 6]
%C [1, 2, 4, 7, 14, 28]
%C [1, 2, 3, 6]
%C ...
%H Michel Lagneau, <a href="/A226871/b226871.txt">Rows n = 1..2223 of irregular triangle, flattened</a>
%e Row 3 = [1, 2, 3, 6] consists of the first 4 divisors of A225110(3) = 18; 1 + 1/2 + 1/3 + 1/6 = 2 is an integer.
%p with(numtheory): print({1}):for n from 1 to 5000 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: print({seq(x[i],i=1..q)}) else fi:od:od:
%Y Cf. A225110, A226853.
%K nonn,tabf
%O 1,3
%A _Michel Lagneau_, Jun 20 2013
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