%I #14 Feb 14 2022 01:25:10
%S 1,2,2,3,5,3,4,9,11,4,5,14,26,19,5,6,20,50,55,30,6,7,27,85,125,105,44,
%T 7,8,35,133,245,280,182,62,8,9,44,196,434,630,560,300,85,9,10,54,276,
%U 714,1260,1428,1056,477,115,10,11,65,375,1110,2310,3192,3030,1905,745,155
%N T(n,k) = k*binomial(n-1,k-1) + Fibonacci(k)*binomial(n-1,k) (1 <= k <= n).
%C Row sums are s(n) = 1, 4, 11, 28, 69, 167, 400, ...
%C Binomial transform of the bidiagonal matrix with (1,2,3...) in the main diagonal and the Fibonacci numbers (1,1,2,3,5,8,...) in the subdiagonal.
%F T(n,2) = A000096(n-1).
%F T(n,3) = A051925(n-1).
%F T(n,4) = A215862(n-3). - _R. J. Mathar_, Aug 10 2013
%F Row sums s(n) = 7*s(n-1) -17*s(n-2) +16*s(n-3) -4*s(n-4) with s(n) = A001787(n+1)/4 +A001906(n-1). - _R. J. Mathar_, Aug 10 2013
%e First few rows of the triangle:
%e 1;
%e 2, 2;
%e 3, 5, 3;
%e 4, 9, 11, 4;
%e 5, 14, 26, 19, 5;
%e 6, 20, 50, 55, 30, 6;
%e 7, 27, 85, 125, 105, 44, 7;
%e 8, 35, 133, 245, 280, 182, 62, 8;
%e ...
%p with(combinat): T:=(n,k)->k*binomial(n-1,k-1)+fibonacci(k)*binomial(n-1,k): for n from 1 to 12 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form
%t Flatten[Table[k Binomial[n-1,k-1]+Fibonacci[k]Binomial[n-1,k],{n,15},{k,n}]] (* _Harvey P. Dale_, Nov 03 2014 *)
%Y Cf. A000096, A051925.
%K nonn,tabl
%O 1,2
%A _Gary W. Adamson_, Nov 20 2006
%E Edited by _N. J. A. Sloane_, Nov 29 2006
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