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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.
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%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