OFFSET
1,3
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).
LINKS
Michel Marcus, Rows n = 1..100 of triangle, flattened
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)
EXAMPLE
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 ].
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 *)
PROG
(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], ", ")))}
(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] ];
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Klaus Brockhaus, following a suggestion of Paul Curtz, Jun 10 2007
STATUS
approved