%I
%S 3,9,15,3,21,0,27,9,33,0,3,39,15,0,45,0,0,51,21,9,57,0,0,3,63,27,0,0,
%T 69,0,15,0,75,33,0,0,81,0,0,9,87,39,21,0,3,93,0,0,0,0,99,45,0,0,0,105,
%U 0,27,15,0,111,51,0,0,0,117,0,0,0,9,123,57,33,0,0,3,129,0,0,21,0,0,135,63,0,0,0,0,141,0,39,0,0,0
%N Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the numbers A016945 interleaved with k1 zeros, and the first element of column k is in row k(k+1)/2.
%C Alternating sum of row n equals 3 times sigma(n), i.e., sum_{k=1..A003056(n))} (1)^(k1)*T(n,k) = 3*A000203(n) = A272027(n).
%C Row n has length A003056(n) hence the first element of column k is in row A000217(k).
%C The number of positive terms in row n is A001227(n).
%C If T(n,k) = 9 then T(n+1,k+1) = 3 is the first element of the column k+1.
%C For more information see A196020.
%F T(n,k) = 3*A196020(n,k) = A196020(n,k) + A236106(n,k).
%e Triangle begins:
%e 3;
%e 9;
%e 15, 3;
%e 21, 0;
%e 27, 9;
%e 33, 0, 3;
%e 39, 15, 0;
%e 45, 0, 0;
%e 51, 21, 9;
%e 57, 0, 0, 3;
%e 63, 27, 0, 0;
%e 69, 0, 15, 0;
%e 75, 33, 0, 0;
%e 81, 0, 0, 9;
%e 87, 39, 21, 0, 3;
%e 93, 0, 0, 0, 0;
%e 99, 45, 0, 0, 0;
%e 105, 0, 27, 15, 0;
%e 111, 51, 0, 0, 0;
%e 117, 0, 0, 0, 9;
%e 123, 57, 33, 0, 0, 3;
%e 129, 0, 0, 21, 0, 0;
%e 135, 63, 0, 0, 0, 0;
%e 141, 0, 39, 0, 0, 0;
%e ...
%e For n = 9 the divisors of 9 are 1, 3, 9, therefore the sum of the divisors of 9 is 1 + 3 + 9 = 13 and 3*13 = 39. On the other hand the 9th row of triangle is 51, 21, 9, therefore the alternating row sum is 51  21 + 9 = 39, equaling 3 times sigma(9).
%Y Column 1 is A016945.
%Y Cf. A000203, A001227, A003056, A074400, A196020, A236106, A237048, A237593, A239050, A239662, A244050, A272027.
%K nonn,tabf
%O 1,1
%A _Omar E. Pol_, Apr 18 2016
