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%I #15 Sep 25 2020 10:38:00
%S 1,4,1,9,4,1,20,9,4,1,40,20,9,4,1,78,40,20,9,4,1,147,78,40,20,9,4,1,
%T 272,147,78,40,20,9,4,1,495,272,147,78,40,20,9,4,1,890,495,272,147,78,
%U 40,20,9,4,1,1584,890,495,272,147,78,40,20,9,4,1,2796,1584,890,495,272
%N Triangle read by rows: T(n,k) = (n+1-k)*Fibonacci(n+2-k), for n>=1, 1<=k<=n.
%C The first column is A023607 (without the leading zero).
%H Harvey P. Dale, <a href="/A104796/b104796.txt">Table of n, a(n) for n = 1..1000</a>
%e Rows 1,2,3,4,5,6 and columns 1,2,3,4,5,6 of the triangle are:
%e 1;
%e 4, 1;
%e 9, 4, 1;
%e 20, 9, 4, 1;
%e 40, 20, 9, 4, 1;
%e 78, 40, 20, 9, 4, 1;
%e ...
%e Row 3 for example is 3*F(4), 2*F(3), 1*F(2) = 3*3, 2*2, 1*1 = 9, 4, 1.
%e Row 4 is 4*F(5), 3*F(4), 2*F(3), 1*F(2) = 4*5, 3*3, 2*2, 1*1 = 20, 9, 4, 1.
%e Reading the rows backwards gives an initial segment of the terms of A023607 (but without the initial zero).
%t Table[(n+1-k)Fibonacci[n+2-k],{n,20},{k,n}]//Flatten (* _Harvey P. Dale_, Sep 24 2020 *)
%t Module[{nn=20,c},c=LinearRecurrence[{2,1,-2,-1},{1,4,9,20},nn];Table[ Reverse[ Take[c,n]],{n,nn}]]//Flatten (* _Harvey P. Dale_, Sep 25 2020 *)
%Y Row sums are in A094584.
%Y Cf. A023607, A104762, A104765, A000045.
%K nonn,tabl
%O 1,2
%A _Gary W. Adamson_, Mar 26 2005
%E Edited by _Ralf Stephan_, Apr 05 2009
%E Entry revised by _N. J. A. Sloane_, Sep 23 2020
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