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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.

%I A347584 #35 Sep 14 2021 03:49:40
%S A347584 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,
%T A347584 18,29,45,62,73,73,62,45,29,47,74,107,135,146,135,107,74,47,76,121,
%U A347584 181,242,281,281,242,181,121,76,123,197,302,423,523,562,523,423,302,197,123
%N A347584 Triangle formed by Pascal's rule, except that the n-th row begins and ends with the n-th Lucas number.
%C A347584 Similar in spirit to the Fibonacci-Pascal triangle A074829, which uses Fibonacci numbers instead of Lucas numbers at the ends of each row.
%C A347584 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.
%H A347584 <a href="/index/Pas#Pascal">Index entries for triangles and arrays related to Pascal's triangle</a>
%F A347584 a(n) = 2*A074829(n+1) - A108617(n).
%e A347584 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.
%e A347584 Triangle begins:
%e A347584                2;
%e A347584              1,  1;
%e A347584            3,  2,  3;
%e A347584          4,  5,  5,  4;
%e A347584        7,  9, 10,  9,  7;
%e A347584     11, 16, 19, 19, 16, 11;
%e A347584   18, 27, 35, 38, 35, 27, 18;
%t A347584 T[n_, 0] := LucasL[n]; T[n_, n_] := LucasL[n];
%t A347584 T[n_, k_] := T[n - 1, k - 1] + T[n - 1, k];
%t A347584 Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten
%Y A347584 Cf. A000032, A027973.
%Y A347584 Cf. A227550, A228196 (general formula).
%Y A347584 Fibonacci borders: A074829, A108617, A316938, A316939.
%K A347584 nonn
%O A347584 0,1
%A A347584 _Noah Carey_ and _Greg Dresden_, Sep 07 2021