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A347584
Triangle formed by Pascal's rule, except that the n-th row begins and ends with the n-th Lucas number.
0
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.
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. A227550, A228196 (general formula).
Fibonacci borders: A074829, A108617, A316938, A316939.
Sequence in context: A049346 A227310 A291905 * A366398 A365932 A240853
KEYWORD
nonn
AUTHOR
Noah Carey and Greg Dresden, Sep 07 2021
STATUS
approved