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 A094439 Triangular array T(n,k)=F(k+4)C(n,k), k=0,1,2,3,...,n; n>=0. 1

%I

%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

%N Triangular array T(n,k)=F(k+4)C(n,k), k=0,1,2,3,...,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).

%e First four rows:

%e 3

%e 3 5

%e 3 10 8

%e 3 15 24 13 sum = 3+15+24+13=55=F(10); alt.sum = 3-15+24-13=-1=-F(-1).

%e T(3,2)=F(5)C(3,2)=5*3=15.

%Y Cf. A094444, A000045.

%K nonn,tabl

%O 1,1

%A _Clark Kimberling_, May 03 2004

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Last modified August 17 11:51 EDT 2019. Contains 326057 sequences. (Running on oeis4.)