A131022 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.

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

Offset: 1

Author

Klaus Brockhaus, following a suggestion of Paul Curtz, Jun 10 2007

Keywords

• nonn
• tabl

Comments

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).

Examples

			First seven rows of T are
[ 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 ].
		

Links

• Michel Marcus, Rows n = 1..100 of triangle, flattened

Crossrefs

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.

Programs

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] ];

Mathematica

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 *)

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], ",")))}

Formula

From Werner Schulte, Jul 22 2017: (Start)

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

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