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Table read by ascending antidiagonals: T(n,j) = Fibonacci(n)*Lucas(n+j), product of the n-th term in the Fibonacci sequence (with F(1)=1 and F(2)=1) and the (n+j)-th term in the Lucas sequence (with L(1)=1 and L(2)=3 and j=0,1,2,...).
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%I #24 Mar 28 2021 00:22:29

%S 1,3,3,8,4,4,21,14,7,7,55,33,22,11,11,144,90,54,36,18,18,377,232,145,

%T 87,58,29,29,987,611,376,235,141,94,47,47,2584,1596,988,608,380,228,

%U 152,76,76,6765,4182,2583,1599,984,615,369,246,123,123

%N Table read by ascending antidiagonals: T(n,j) = Fibonacci(n)*Lucas(n+j), product of the n-th term in the Fibonacci sequence (with F(1)=1 and F(2)=1) and the (n+j)-th term in the Lucas sequence (with L(1)=1 and L(2)=3 and j=0,1,2,...).

%C j is the offset when combining terms from the two initial sequences.

%F For phi=(1+sqrt(5))/2 and tau=(1-sqrt(5))/2:

%F T(n,j) = Fibonacci(n)*Lucas(n+j).

%F T(n,j) = (phi^n - tau^n)*(phi^(n+j) + tau^(n+j))/sqrt(5).

%F T(n,j) = Fibonacci(2n+j) - (-1)^n*Fibonacci(j).

%F Lim_{n, j -> oo} T(n+1,j)/T(n,j) = phi^2 (A104457).

%F Lim_{n, j -> oo} T(n,j+1)/T(n,j) = phi (A001622).

%e T(4,3) = Fibonacci(4)*Lucas(4+3) = 3*29 = 87.

%e Square array showing T(n,j) begins:

%e j=0 j=1 j=2 j=3 j=4 ..

%e n=1 1 3 4 7 11 ..

%e n=2 3 4 7 11 18 ..

%e n=3 8 14 22 36 58 ..

%e n=4 21 33 54 87 141 ..

%e ... .. .. .. .. .. ..

%o (PARI) T(n,j) = fibonacci(2*n+j) - (-1)^n*fibonacci(j);

%o matrix(7,7,n,k, T(n,k-1)) \\ _Michel Marcus_, Mar 02 2021

%Y For j=0 the resulting sequence is used as input in A341414.

%Y Cf. A000045, A000032, A001622, A094214, A104457.

%K easy,nonn,tabl

%O 1,2

%A _Jens Rasmussen_, Feb 16 2021