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%I #31 Jun 13 2021 03:25:16
%S 1,3,4,5,2,6,7,5,8,9,6,10,3,11,7,12,13,8,7,14,15,9,16,8,17,10,4,18,19,
%T 11,9,20,21,12,9,22,10,23,13,24,25,14,11,10,26,5,27,15,28,12,29,16,11,
%U 30,31,17,13,11,32,33,18,12,34,14,35,19,36,12,37,20,15,13,6,38,39,21,40,16,41,22,14,13,42,43,23,17,13
%N Irregular triangle read by rows, n >= 1, 1 <= k <= A038548(n), in which T(n,k) is the sum of the k-th pair of conjugate divisors of n, or T(n,k) is the central divisor of n if such a pair does not exist.
%H Omar E. Pol, <a href="http://www.polprimos.com/imagenespub/poldiv05.jpg">Illustration of the divisors of the first 12 positive integers</a>
%H <a href="/index/Si#SIGMAN">Index entries for sequences related to sigma(n)</a>
%e Triangle begins:
%e 1;
%e 3;
%e 4;
%e 5, 2;
%e 6;
%e 7, 5;
%e 8;
%e 9, 6;
%e 10, 3;
%e 11, 7;
%e 12;
%e 13, 8, 7;
%e ...
%e For n = 9 the divisors of 9 are [1, 3, 9]. There is only one pair of conjugate divisors: [1, 9], and the central divisor is 3, so the 9th row of the triangle is [10, 3].
%e For n = 12 the divisors of 12 are [1, 2, 3, 4, 6, 12]. There are three pairs of conjugate divisors, they are [1, 12], [2, 6], [3, 4], so the 12th row of the triangle is [13, 8, 7].
%Y Row sums give A000203.
%Y Row lengths give A038548.
%Y Right border gives A207376.
%Y Column 1 is A065475.
%Y Cf. A027750, A210959, A228814.
%K nonn,tabf
%O 1,2
%A _Omar E. Pol_, Apr 21 2016
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