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%I #15 Apr 11 2016 23:14:47
%S 1,6,3,17,13,4,38,31,19,7,80,69,48,32,11,158,140,107,79,51,18,303,274,
%T 220,176,127,83,29,566,519,432,360,283,206,134,47,1039,963,822,706,
%U 580,459,333,217,76,1880,1757,1529,1341,1138,940,742,539,351,123,3364,3165,2796,2492,2163,1844,1520,1201
%N Triangle A067979 with rows read backwards.
%C The column m (without leading 0's) gives the convolution of Lucas numbers {L(n+1) := A000204(n+1)}, n>=0, with those with m-shifted index: a(n+m,m)=sum(L(k+1)*L(m+n+1-k),k=0..n), n>=0,m=0,1,...
%C The columns give A004799(n-1), A067980-7 for m= 0..8, respectively. Row sums give A067989.
%C The row polynomials p(n,x) := sum(a(n,m)*x^m,m=0..n) are generated by A(z)*(A(z)-x*A(x*z))/(1-x), with A(x) := (1+2*x)/(1-x-x^2) (g.f. for Lucas {L(n+1)}).
%H Michael De Vlieger, <a href="/A067990/b067990.txt">Table of n, a(n) for n = 0..10000</a>
%F a(n, m)=A067330(n, n-m), n>=m>=0, else 0.
%F a(n, m)=(n-m+1)*L(m+1)*F(n-m)+((n-m+1)*L(m+1)+(n-m)*L(m))*F(n-m+1), n>=m>=0, else 0; with F(n) := A000045(n)(Fibonacci) and L(n) := A000032(n) (Lucas).
%F G.f. for column m=0, 1, ...: (x^m)*(L(m+1)+L(m)*x)*(1+2*x)/(1-x-x^2)^2.
%F a(n, m) = -(-1)^m*F(n-2*m+1)-m*L(n+2)+n*L(n+2)+F(n+3), with F(-n) = (-1)^(n+1)*F(n), hence a(n, m) = -5*A067418(n, m)+2*(n-m+1)*L(n+2), n>=m>=0. - _Ehren Metcalfe_, Apr 11 2016
%e {1}; {6,3}; {17,13,4}; {38,31,19,7}; ...; p(2,x)=17+13*x+4*x^2.
%t Reverse /@ Table[Sum[LucasL[k + 1] LucasL[n - k + 1], {k, 0, m}], {n, 0, 11}, {m, 0, n}] // Flatten (* _Michael De Vlieger_, Apr 11 2016 *)
%K nonn,easy,tabl
%O 0,2
%A _Wolfdieter Lang_, Feb 15 2002
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