A272025 - OEIS

%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