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Triangular array read by rows: T(n,k) = Fibonacci(k)*C(n,k), k = 1...n; n>=1.
11

%I #11 Sep 08 2022 08:45:13

%S 1,2,1,3,3,2,4,6,8,3,5,10,20,15,5,6,15,40,45,30,8,7,21,70,105,105,56,

%T 13,8,28,112,210,280,224,104,21,9,36,168,378,630,672,468,189,34,10,45,

%U 240,630,1260,1680,1560,945,340,55,11,55,330,990,2310,3696,4290,3465,1870,605,89

%N Triangular array read by rows: T(n,k) = Fibonacci(k)*C(n,k), k = 1...n; n>=1.

%C Let F(n) denote the n-th Fibonacci number (A000045). Then n-th row sum of T is F(2n) and n-th alternating row sum is F(n).

%H G. C. Greubel, <a href="/A094435/b094435.txt">Rows n = 1..100 of triangle, flattened</a>

%F From _G. C. Greubel_, Oct 30 2019: (Start)

%F T(n, k) = binomial(n, k)*Fibonacci(k).

%F Sum_{k=1..n} binomial(n,k)*Fibonacci(k) = Fibonacci(2*n).

%F Sum_{k=1..n} (-1)^(k-1)*binomial(n,k)*Fibonacci(k) = Fibonacci(n). (End)

%e First few rows:

%e 1;

%e 2 1;

%e 3 3 2;

%e 4 6 8 3;

%e 5, 10, 20, 15, 5;

%e 6, 15, 40, 45, 30, 8;

%p with(combinat); seq(seq(binomial(n,k)*fibonacci(k), k=1..n), n=1..12); # _G. C. Greubel_, Oct 30 2019

%t Table[Fibonacci[k]*Binomial[n, k], {n, 12}, {k, n}]//Flatten (* _G. C. Greubel_, Oct 30 2019 *)

%o (PARI) T(n,k) = binomial(n,k)*fibonacci(k);

%o for(n=1,12, for(k=1,n, print1(T(n,k), ", "))) \\ _G. C. Greubel_, Oct 30 2019

%o (Magma) [Binomial(n,k)*Fibonacci(k): k in [1..n], n in [1..12]]; // _G. C. Greubel_, Oct 30 2019

%o (Sage) [[binomial(n,k)*fibonacci(k) for k in (1..n)] for n in (1..12)] # _G. C. Greubel_, Oct 30 2019

%o (GAP) Flat(List([1..12], n-> List([1..n], k-> Binomial(n,k)*Fibonacci(k) ))); # _G. C. Greubel_, Oct 30 2019

%Y Cf. A000045.

%Y Cf. A094436, A094437, A094438, A094439, A094440, A094441, A094442, A094443, A094444.

%K nonn,tabl

%O 1,2

%A _Clark Kimberling_, May 03 2004