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%I #10 Feb 22 2013 14:40:24
%S 1,1,1,1,3,2,1,6,8,3,1,10,21,17,5,1,15,45,58,35,8,1,21,85,154,144,68,
%T 13,1,28,147,350,452,330,129,21,1,36,238,714,1195,1198,719,239,34,1,
%U 45,366,1344,2799,3611,2959,1506,436,55
%N Triangle T(n,k), read by rows, given by (1, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.
%C T(n,n) = Fibonacci(n+1) = A000045(n+1).
%C A202390 is jointly generated with A208340 as an array of coefficients of polynomials v(n,x): initially, u(1,x)=v(1,x)=1; for n>1, u(n,x)=u(n-1,x)+x*v(n-1)x and v(n,x)=(x+1)*u(n-1,x)+(x+1)v(n-1,x). The alternating row sums of A202390, and also A208340, are 0 except for the first one. See the Mathematica section. [From Clark Kimberling, Feb 27 2012]
%F T(n,k) = 2*T(n-1,k) + T(n-1,k-1) + T(n-2,k-2) - T(n-2,k) with T(0,0) = T(1,0) = T(1,1) = 1 and T(n,k) = 0 if k<0 or if n<k.
%F G.f.: (1-x)/(1-(2+y)*x+(1-y^2)*x^2).
%F Sum_{k, 0<=k<=n} T(n,k)*x^k = (-1)^n*A108411(n), A000007(n), A000012(n), A025192(n), A122558(n) for x = -2, -1, 0, 1, 2 respectively.
%e Triangle begins :
%e 1
%e 1, 1
%e 1, 3, 2
%e 1, 6, 8, 3
%e 1, 10, 21, 17, 5
%e 1, 15, 45, 58, 35, 8
%e 1, 21, 85, 154, 144, 68, 13
%e 1, 28, 147, 350, 452, 330, 129, 21
%t u[1, x_] := 1; v[1, x_] := 1; z = 13;
%t u[n_, x_] := u[n - 1, x] + x*v[n - 1, x];
%t v[n_, x_] := (x + 1)*u[n - 1, x] + (x + 1)*v[n - 1, x];
%t Table[Expand[u[n, x]], {n, 1, z/2}]
%t Table[Expand[v[n, x]], {n, 1, z/2}]
%t cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
%t TableForm[cu]
%t Flatten[%] (* A202390 *)
%t Table[Expand[v[n, x]], {n, 1, z}]
%t cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
%t TableForm[cv]
%t Flatten[%] (* A208340 *)
%t Table[u[n, x] /. x -> 1, {n, 1, z}] (*row sums*)
%t Table[u[n, x] /. x -> -1, {n, 1, z}] (*alt. row sums*)
%Y Cf. A000012, A000217, A051744, A000045, A123585, A208340.
%K nonn,tabl
%O 0,5
%A _Philippe Deléham_, Dec 18 2011
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