A347584 Triangle formed by Pascal's rule, except that the n-th row begins and ends with the n-th Lucas number.

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, 29, 45, 62, 73, 73, 62, 45, 29, 47, 74, 107, 135, 146, 135, 107, 74, 47, 76, 121, 181, 242, 281, 281, 242, 181, 121, 76, 123, 197, 302, 423, 523, 562, 523, 423, 302, 197, 123

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

Table of n, a(n) for n=0..65.

Index entries for triangles and arrays related to Pascal's triangle

Formula

a(n) = 2*A074829(n+1) - A108617(n).

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

Cf. A000032, A027973.

Cf. A227550, A228196 (general formula).

Fibonacci borders: A074829, A108617, A316938, A316939.

Sequence in context: A049346 A227310 A291905 * A366398 A365932 A240853

Adjacent sequences: A347581 A347582 A347583 * A347585 A347586 A347587

Keyword

nonn

Author

Noah Carey and Greg Dresden, Sep 07 2021

Status

approved