%I #13 Apr 02 2017 01:56:23
%S 1,3,1,4,2,7,3,1,6,4,12,5,2,2,8,6,15,7,3,1,13,8,4,18,9,4,0,12,10,28,
%T 11,5,4,3,1,14,12,24,13,6,2,24,14,8,8,31,15,7,3,1,18,16,39,17,8,10,4,
%U 4,20,18,42,19,11,4,5,1,32,20,12,8,36,21,10,6,24,22,60,23,11,10,6,5,2,2,31,24,16,42,25,12,8
%N Irregular triangle read by rows: T(n,k) = sum of the elements of the k-th row of the absolute difference table of the divisors of n.
%C Row 2^k gives the first k+1 positive terms of A000225 in decreasing order, k >= 0.
%C If n is prime then row n contains only two terms: n+1 and n-1.
%C Note that this sequence is not the absolute values of A273261.
%C First differs from A273261 at a(41).
%e Triangle begins:
%e 1;
%e 3, 1;
%e 4, 2;
%e 7, 3, 1;
%e 6, 4;
%e 12, 5, 2, 2;
%e 8, 6;
%e 15, 7, 3, 1;
%e 13, 8, 4;
%e 18, 9, 4, 0;
%e 12, 10;
%e 28, 11, 5, 4, 3, 1;
%e 14, 12;
%e 24, 13, 6, 2;
%e 24, 14, 8, 8;
%e 31, 15, 7, 3, 1;
%e 18, 16;
%e 39, 17, 8, 10, 4, 4;
%e 20, 18;
%e 42, 19, 11, 4, 5, 1;
%e 32, 20, 12, 8;
%e 36, 21, 10, 6;
%e 24, 22;
%e 60, 23, 11, 10, 6, 5, 2, 2;
%e 31, 24, 16;
%e 42, 25, 12, 8;
%e ...
%e For n = 18 the divisors of 18 are 1, 2, 3, 6, 9, 18, and the absolute difference triangle of the divisors is
%e 1, 2, 3, 6, 9, 18;
%e 1, 1, 3, 3, 9;
%e 0, 2, 0, 6;
%e 2, 2, 6;
%e 0, 4;
%e 4;
%e The row sums give [39, 17, 8, 10, 4, 4] which is also the 18th row of the irregular triangle.
%t Map[Total, #, {2}] &@ Table[NestWhileList[Abs@ Differences@ # &, #, Length@ # > 1 &] &@ Divisors@ n, {n, 26}] // Flatten (* _Michael De Vlieger_, Jun 27 2016 *)
%Y Row lengths give A000005. Column 1 is A000203. Right border gives A187203. Row sums give A187215.
%Y Cf. A000225, A273104, A273261, A274532, A274533.
%K nonn,tabf
%O 1,2
%A _Omar E. Pol_, Jun 27 2016