OFFSET
0,1
COMMENTS
Similar in spirit to the Fibonacci-Pascal triangle A074829, which uses Fibonacci numbers instead of Lucas numbers at the ends of each row.
If we consider the top of the triangle to be the 0th row, then the sum of terms in n-th row is 2*(2^(n+1) - Lucas(n+1)). This sum also equals 2*A027973(n-1) for n>0.
LINKS
EXAMPLE
The first two Lucas numbers (for n=0 and n=1) are 2 and 1, so the first two rows (again, for n=0 and n=1) of the triangle are 2 and 1, 1 respectively.
Triangle begins:
2;
1, 1;
3, 2, 3;
4, 5, 5, 4;
7, 9, 10, 9, 7;
11, 16, 19, 19, 16, 11;
18, 27, 35, 38, 35, 27, 18;
MATHEMATICA
T[n_, 0] := LucasL[n]; T[n_, n_] := LucasL[n];
T[n_, k_] := T[n - 1, k - 1] + T[n - 1, k];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten
CROSSREFS
KEYWORD
nonn
AUTHOR
Noah Carey and Greg Dresden, Sep 07 2021
STATUS
approved