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%I #11 Oct 30 2019 16:50:35
%S 1,1,1,1,2,1,1,2,1,4,1,1,1,3,1,6,1,1,2,4,1,2,6,1,1,3,5,1,10,1,1,1,1,2,
%T 6,1,12,1,1,5,7,1,2,2,10,1,1,2,4,8,1,16,1,1,1,3,3,9,1,18,1,1,2,1,5,10,
%U 1,2,4,14,1,1,9,11,1,22,1,1,1,1,2,2,4,12,1,4,20,1,1,11,13,1,2,6,18,1,1,2,3
%N Irregular array: the sum of the first m terms of row n is the m-th positive divisor of n.
%C Row n contains A000005(n) terms.
%e The positive divisors of 20 are 1,2,4,5,10,20. Row 20 of the array is 1,1,2,1,5,10. So we have 1=1; 1+1=2; 1+1+2=4; 1+1+2+1=5; 1+1+2+1+5=10; 1+1+2+1+5+10=20.
%e The first 10 rows:
%e 1;
%e 1,1;
%e 1,2;
%e 1,1,2;
%e 1,4;
%e 1,1,1,3
%e 1,6;
%e 1,1,2,4;
%e 1,2,6;
%e 1,1,3,5
%t f[n_] := Block[{d},d = Divisors[n]; d - Prepend[Most[d], 0]]; Flatten[Array[f, 40]] (* _Ray Chandler_, Nov 01 2007 *)
%Y Cf. A027750, A000005.
%K nonn,tabf
%O 1,5
%A _Leroy Quet_, Oct 30 2007
%E Extended by _Ray Chandler_, Nov 01 2007
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