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Irregular triangle read by rows where n-th row (of A061395(n) terms, for n>=2) is such that n = Product_{j=1..A061395(n)} prime(j)^(Sum_{k=1..j} T(n,k)). Row 1 is {0}.
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%I #16 Aug 10 2021 11:11:56

%S 0,1,0,1,2,0,0,1,1,0,0,0,0,1,3,0,2,1,-1,1,0,0,0,0,1,2,-1,0,0,0,0,0,1,

%T 1,-1,0,1,0,1,0,4,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,0,1,2,-2,1,0,1,-1,1,1,

%U -1,0,0,1,0,0,0,0,0,0,0,0,1,3,-2,0,0,2

%N Irregular triangle read by rows where n-th row (of A061395(n) terms, for n>=2) is such that n = Product_{j=1..A061395(n)} prime(j)^(Sum_{k=1..j} T(n,k)). Row 1 is {0}.

%C The rows of this triangle also give all the ordered ways that a finite number of integers can be arranged so that their partial sums, from left to right, are all nonnegative and their total sum is positive.

%e Triangle begins:

%e 0;

%e 1;

%e 0, 1;

%e 2;

%e 0, 0, 1;

%e 1, 0;

%e 0, 0, 0, 1;

%e 3;

%e ...

%e Row 20 is {2, -2, 1}. So 20 = prime(1)^T(20,1) * prime(2)^(T(20,1) + T(20,2)) * prime(3)^(T(20,1) + T(20,2) + T(20,3)) = 2^2 * 3^(2 - 2) * 5^(2 - 2 + 1) = 2^2 * 3^0 * 5^1.

%Y Cf. A061395, A067255, A134364.

%K sign,tabf

%O 1,5

%A _Leroy Quet_, Oct 22 2007