%I #14 Sep 08 2022 08:45:30

%S 1,1,2,1,2,4,0,1,3,7,0,0,1,4,11,0,0,0,1,5,16,1,1,1,1,2,7,23,1,2,3,4,5,

%T 7,14,37,1,2,4,7,11,16,23,37,74,0,1,3,7,14,25,41,64,101,175,0,0,1,4,

%U 11,25,50,91,155,256,431,0,0,0,1,5,16,41,91,182,337,593,1024

%N Triangular array T read by rows: T(j,1) = 1 for ((j-1) mod 6) < 3, else 0; T(j,k) = T(j-1,k-1) + T(j,k-1) for 2 <= k <= j.

%C All columns are periodic with period length 6. The (3+6*i)-th row equals the first (3+6*i) terms of main diagonal (i >= 0).

%H Michel Marcus, <a href="/A131022/b131022.txt">Rows n = 1..100 of triangle, flattened</a>

%F From _Werner Schulte_, Jul 22 2017: (Start)

%F T(n,k) = 2^(k-2) + 2*sqrt(3)^(k-3) * sin(Pi/6*(2*n-k)) for 1 < k <= n, and

%F T(n,1) = 1 - floor((n-1)/3) mod 2 for n >= 1. (End)

%e First seven rows of T are

%e [ 1 ]

%e [ 1, 2 ]

%e [ 1, 2, 4 ]

%e [ 0, 1, 3, 7 ]

%e [ 0, 0, 1, 4, 11 ]

%e [ 0, 0, 0, 1, 5, 16 ]

%e [ 1, 1, 1, 1, 2, 7, 23 ].

%t T[j_, 1] := If[Mod[j-1, 6]<3, 1, 0]; T[j_, k_] := T[j, k] = T[j-1, k-1]+T[j, k-1]; Table[T[j, k], {j, 1, 13}, {k, 1, j}] // Flatten (* _Jean-François Alcover_, Mar 06 2014 *)

%o (PARI) {m=13; M=matrix(m, m); for(j=1, m, M[j, 1]=if((j-1)%6<3, 1, 0)); for(k=2, m, for(j=k, m, M[j, k]=M[j-1, k-1]+M[j, k-1])); for(j=1, m, for(k=1, j, print1(M[j, k], ",")))}

%o (Magma) m:=13; M:=ZeroMatrix(IntegerRing(), m, m); for j:=1 to m do if (j-1) mod 6 lt 3 then M[j, 1]:=1; end if; end for; for k:=2 to m do for j:=k to m do M[j, k]:=M[j-1, k-1]+M[j, k-1]; end for; end for; &cat[ [ M[j, k]: k in [1..j] ]: j in [1..m] ];

%Y Cf. A129339 (main diagonal of T), A131023 (first subdiagonal of T), A131024 (row sums of T), A131025 (antidiagonal sums of T). First through sixth column of T are in A088911, A131026, A131027, A131028, A131029, A131030 resp.

%K nonn,tabl

%O 1,3

%A _Klaus Brockhaus_, following a suggestion of _Paul Curtz_, Jun 10 2007