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%I #27 Feb 26 2023 19:35:20
%S 1,2,1,2,2,1,1,2,2,1,0,1,2,2,1,0,0,1,2,2,1,1,0,0,1,2,2,1,2,1,0,0,1,2,
%T 2,1,2,2,1,0,0,1,2,2,1,1,2,2,1,0,0,1,2,2,1,0,1,2,2,1,0,0,1,2,2,1,0,0,
%U 1,2,2,1,0,0,1,2,2,1
%N Triangle read by rows: partial sums from the right of an A010892 subsequences decrescendo triangle.
%C n-th row = (n+1) terms of an infinitely periodic cycle: (..., 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1), shifting to the right one place for the next row.
%C Construct an A010892 decrescendo triangle: (1; 1,1; 0,1,1; -1,0,1,1; ...) and take partial sums starting from the right.
%F T(n, k) = 1 + A010892(n-k-1), with 0 <= k <= n. - _Stefano Spezia_, Feb 11 2023
%e First few rows of the triangle:
%e 1;
%e 2, 1;
%e 2, 2, 1;
%e 1, 2, 2, 1;
%e 0, 1, 2, 2, 1;
%e 0, 0, 1, 2, 2, 1;
%e 1, 0, 0, 1, 2, 2, 1;
%e 2, 1, 0, 0, 1, 2, 2, 1;
%e 2, 2, 1, 0, 0, 1, 2, 2, 1;
%e 1, 2, 2, 1, 0, 0, 1, 2, 2, 1;
%e 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1;
%e ...
%e Row 3 = (1, 2, 2, 1) = partial sums of (-1, 0, 1, 1).
%t A010892[n_]:={1, 1, 0, -1, -1,0}[[Mod[n, 6]+1]]; T[n_,k_]:=1+A010892[n-k-1]; Table[T[n,k], {n,0, 11},{k,0,n}]//Flatten (* _Stefano Spezia_, Feb 11 2023 *)
%Y Cf. A010892, A077859 (row sums), A164965 (1st column).
%K nonn,tabl
%O 0,2
%A _Gary W. Adamson_, Sep 10 2008
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