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Triangular array T(n,k) = Fibonacci(k+4)*C(n,k), k=0..n, n>=0.
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%I #12 Sep 08 2022 08:45:13

%S 3,3,5,3,10,8,3,15,24,13,3,20,48,52,21,3,25,80,130,105,34,3,30,120,

%T 260,315,204,55,3,35,168,455,735,714,385,89,3,40,224,728,1470,1904,

%U 1540,712,144,3,45,288,1092,2646,4284,4620,3204,1296,233,3,50,360,1560,4410,8568,11550,10680,6480,2330,377

%N Triangular array T(n,k) = Fibonacci(k+4)*C(n,k), k=0..n, n>=0.

%C Let F(n) denote the n-th Fibonacci number (A000045). Then n-th row sum of T is F(2n+4) and n-th alternating row sum is -F(n-4).

%H G. C. Greubel, <a href="/A094439/b094439.txt">Rows n = 0..100 of triangle, flattened</a>

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

%F T(n, k) = binomial(n,k)*Fibonacci(k+4).

%F Sum_{k=0..n} T(n,k) = Fibonacci(2*n+4).

%F Sum_{k=0..n} (-1)^(k+1) * T(n,k) = Fibonacci(n-4). (End)

%e First few rows:

%e 3;

%e 3, 5;

%e 3, 10, 8;

%e 3, 15, 24, 13;

%e 3, 20, 48, 52, 21;

%e 3, 25, 80, 130, 105, 34;

%p with(combinat); seq(seq(fibonacci(k+4)*binomial(n,k), k=0..n), n=0..12); # _G. C. Greubel_, Oct 30 2019

%t Table[Fibonacci[k+4]*Binomial[n,k], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, Oct 30 2019 *)

%o (PARI) T(n,k) = binomial(n,k)*fibonacci(k+4);

%o for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ _G. C. Greubel_, Oct 30 2019

%o (Magma) [Binomial(n,k)*Fibonacci(k+4): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Oct 30 2019

%o (Sage) [[binomial(n,k)*fibonacci(k+4) for k in (0..n)] for n in (0..12)] # _G. C. Greubel_, Oct 30 2019

%o (GAP) Flat(List([0..12], n-> List([0..n], k-> Binomial(n,k)* Fibonacci(k+4) ))); # _G. C. Greubel_, Oct 30 2019

%Y Cf. A000045, A007318, A094435, A094436, A094437, A094438, A094440, A094441, A094442, A094443, A094444.

%K nonn,tabl

%O 0,1

%A _Clark Kimberling_, May 03 2004