%I
%S 4,12,20,4,28,0,36,12,44,0,4,52,20,0,60,0,0,68,28,12,76,0,0,4,84,36,0,
%T 0,92,0,20,0,100,44,0,0,108,0,0,12,116,52,28,0,4,124,0,0,0,0,132,60,0,
%U 0,0,140,0,36,20,0,148,68,0,0,0,156,0,0,0,12,164,76,44,0,0,4,172,0,0,28,0,0,180,84,0,0,0,0,188,0,52,0,0,0
%N Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the numbers A017113 interleaved with k1 zeros, and the first element of column k is in row k(k+1)/2.
%C Gives an identity for A239050. Alternating sum of row n equals A239050(n), i.e., sum_{k=1..A003056(n))} (1)^(k1)*T(n,k) = 4*A000203(n) = 2*A074400(n) = A239050(n).
%C Row n has length A003056(n) hence the first element of column k is in row A000217(k).
%C Note that if T(n,k) = 12 then T(n+1,k+1) = 4, the first element of the column k+1.
%C The number of positive terms in row n is A001227(n).
%C For more information see A196020.
%C Column 1 is A017113.  _Omar E. Pol_, Apr 17 2016
%F T(n,k) = 2*A236106(n,k) = 4*A196020(n,k).
%e Triangle begins:
%e 4;
%e 12;
%e 20, 4;
%e 28, 0;
%e 36, 12;
%e 44, 0, 4;
%e 52, 20, 0;
%e 60, 0, 0;
%e 68, 28, 12;
%e 76, 0, 0, 4;
%e 84, 36, 0, 0;
%e 92, 0, 20, 0;
%e 100, 44, 0, 0;
%e 108, 0, 0, 12;
%e 116, 52, 28, 0, 4;
%e 124, 0, 0, 0, 0;
%e 132, 60, 0, 0, 0;
%e 140, 0, 36, 20, 0;
%e 148, 68, 0, 0, 0;
%e 156, 0, 0, 0, 12;
%e 164, 76, 44, 0, 0, 4;
%e 172, 0, 0, 28, 0, 0;
%e 180, 84, 0, 0, 0, 0;
%e 188, 0, 52, 0, 0, 0;
%e ...
%e For n = 9, the 9th row of triangle is [68, 28, 12], therefore the alternating row sum is 68  28 + 12 = 52. On the other hand we have that 4*A000203(9) = 2*A074400(9) = A239050(9) = 4*13 = 2*26 = 52, equaling the alternating sum of the 9th row of the triangle.
%Y Cf. A000203, A000217, A003056, A017113, A074400, A196020, A211343, A235791, A236104, A236106, A236112, A237048, A237591, A239050, A239446.
%K nonn,tabf
%O 1,1
%A _Omar E. Pol_, Mar 30 2014
