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%I #14 Jun 06 2021 09:05:20
%S 1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,1,0,0,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,
%T 1,1,1,1,0,1,0,1,0,1,1,1,1,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,1,1,1,1,1,
%U 1,1,0,0,0,0,1,1,1,1,1,1,0,1,1,0,0,0,1,1,0,1,1,1,1,0,0,1,1,0,0,1,1,0,0,1,1
%N A generalized Pascal triangle modulo 2.
%C Row sums are A114212. Diagonal sums are A114214.
%C Row sums of inverse are 0^n (conjecture).
%H Jeffrey Shallit and Lukas Spiegelhofer, <a href="https://arxiv.org/abs/1710.06203">Continuants, run lengths, and Barry's modified Pascal triangle</a>, arXiv:1710.06203 [math.CO], 2017.
%F T(n, k) = (Sum_{j=0..n-k} C(k, j)*C(n-k, j)*(1+(-1)^j)/2) mod 2.
%e Triangle begins
%e 1;
%e 1, 1;
%e 1, 1, 1;
%e 1, 1, 1, 1;
%e 1, 1, 0, 1, 1;
%e 1, 1, 0, 0, 1, 1;
%e 1, 1, 1, 0, 1, 1, 1;
%e 1, 1, 1, 1, 1, 1, 1, 1;
%e 1, 1, 0, 1, 0, 1, 0, 1, 1;
%e 1, 1, 0, 0, 0, 0, 0, 0, 1, 1;
%e 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1;
%o (PARI) T(n,k) = sum(j=0, n-k, binomial(k, j)*binomial(n-k, j)*(1+(-1)^j)/2) % 2; \\ _Michel Marcus_, Jun 06 2021
%Y Cf. A114212, A114214.
%K easy,nonn,tabl
%O 0,1
%A _Paul Barry_, Nov 17 2005
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