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%I #18 Aug 31 2021 02:43:08
%S 1,2,2,3,3,3,4,4,5,5,4,5,6,6,7,7,4,6,8,8,7,9,9,4,7,10,10,11,11,4,6,8,
%T 10,12,12,13,13,8,9,14,14,11,13,15,15,5,8,12,16,16,17,17,8,11,12,15,
%U 18,18,19,19,7,10,10,15,20,20,13,17,21,21,16,13,22,22,23,23,6,7,10,12,16,20,24,24,21,25,25
%N Irregular triangle read by rows: T(n,k) = sum of the elements of the k-th column of the absolute difference table of the divisors of n.
%C If n is prime then row n is [n, n].
%C It appears that the last two terms of the n-th row are [n, n], n > 1.
%C Note that this sequence is not the absolute values of A273263.
%C First differs from A273263 at a(38).
%e Triangle begins:
%e 1;
%e 2, 2;
%e 3, 3;
%e 3, 4, 4;
%e 5, 5;
%e 4, 5, 6, 6;
%e 7, 7;
%e 4, 6, 8, 8;
%e 7, 9, 9;
%e 4, 7, 10, 10;
%e 11, 11;
%e 4, 6, 8, 10, 12, 12;
%e 13, 13;
%e 8, 9, 14, 14;
%e 11, 13, 15, 15;
%e 5, 8, 12, 16, 16;
%e 17, 17;
%e 8, 11, 12, 15, 18, 18;
%e 19, 19;
%e 7, 10, 10, 15, 20, 20;
%e 13, 17, 21, 21;
%e 16, 13, 22, 22;
%e 23, 23;
%e 6, 7, 10, 12, 16, 20, 24, 24;
%e 21, 25, 25;
%e 20, 15, 26, 26;
%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 column sums give [8, 11, 12, 15, 18, 18] which is also the 18th row of the irregular triangle.
%t Table[Total /@ Table[#[[m - k + 1, -k]], {m, Length@ #, 1, -1}, {k, m}] &@ NestWhileList[Abs@ Differences@ # &, Divisors@ n, Length@ # > 1 &], {n, 25}] // Flatten (* _Michael De Vlieger_, Jun 29 2016 *)
%Y Row lengths give A000005. Right border gives A000027. Row sums give A187215.
%Y Cf. A187203, A272121, A273132, A273104, A273137, A273263, A274531, A274532.
%K nonn,tabf
%O 1,2
%A _Omar E. Pol_, Jun 29 2016
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