Search: id:A108299 Results 1-1 of 1 results found. %I A108299 %S A108299 1,1,1,1,1,1,1,1,2,1,1,1,3,2,1,1,1,4,3,3,1,1,1,5,4,6,3,1,1,1,6,5,10,6, 4, %T A108299 1,1,1,7,6,15,10,10,4,1,1,1,8,7,21,15,20,10,5,1,1,1,9,8,28,21,35,20,15, %U A108299 5,1,1,1,10,9,36,28,56,35,35,15,6,1,1,1,11,10,45,36,84,56,70 %V A108299 1,1,-1,1,-1,-1,1,-1,-2,1,1,-1,-3,2,1,1,-1,-4,3,3,-1,1,-1,-5,4,6,-3,-1, 1,-1,-6,5,10,-6, %W A108299 -4,1,1,-1,-7,6,15,-10,-10,4,1,1,-1,-8,7,21,-15,-20,10,5,-1,1,-1,-9,8, 28,-21,-35,20,15, %X A108299 -5,-1,1,-1,-10,9,36,-28,-56,35,35,-15,-6,1,1,-1,-11,10,45,-36,-84,56, 70 %N A108299 Triangle read by rows, 0 <= k <= n: T(n,k)=binomial(n-[(k+1)/2],[k/2])*(-1)^[(k+1)/ 2]. %C A108299 Matrix inverse of A124645 . %C A108299 Let L(n,x) = Sum(T(n,k)*x^(n-k): 0<=k<=n) and Pi=3.14...: %C A108299 L(n,x) = Prod(x - 2*cos((2*k-1)*Pi/(2*n+1)): 1<=k<=n); %C A108299 Sum(T(n,k): 0<=k<=n) = L(n,1) = A010892(n+1); %C A108299 Sum(abs(T(n,k)): 0<=k<=n) = A000045(n+2); %C A108299 abs(T(n,k))=A065941(n,k), T(n,k)=A065941(n,k)*A087960(k); %C A108299 T(2*n,k) + T(2*n+1,k+1) = 0 for 0<=k<=2*n; %C A108299 T(n,0)=A000012(n)=1; T(n,1)=-1 for n>0; %C A108299 T(n,2)=-(n-1) for n>1; T(n,3)=A000027(n)=n for n>2; %C A108299 T(n,4)=A000217(n-3) for n>3; T(n,5)=-A000217(n-4) for n>4; %C A108299 T(n,6)=-A000292(n-5) for n>5; T(n,7)=A000292(n-6) for n>6; %C A108299 T(n,n-3)=A058187(n-3)*(-1)^[n/2] for n>2; %C A108299 T(n,n-2)=A008805(n-2)*(-1)^[(n+1)/2] for n>1; %C A108299 T(n,n-1)=A008619(n-1)*(-1)^[n/2] for n>0; %C A108299 T(n,n) = L(n,0) = (-1)^[(n+1)/2]; %C A108299 L(n,1) = A010892(n+1); L(n,-1) = A061347(n+2); %C A108299 L(n,2) = 1; L(n,-2) = A005408(n)*(-1)^n; %C A108299 L(n,3) = A001519(n); L(n,-3) = A002878(n)*(-1)^n; %C A108299 L(n,4) = A001835(n+1); L(n,-4) = A001834(n)*(-1)^n; %C A108299 L(n,5) = A004253(n); L(n,-5) = A030221(n)*(-1)^n; %C A108299 L(n,6) = A001653(n); L(n,-6) = A002315(n)*(-1)^n; %C A108299 L(n,7) = A049685(n); L(n,-7) = A033890(n)*(-1)^n; %C A108299 L(n,8) = A070997(n); L(n,-8) = A057080(n)*(-1)^n; %C A108299 L(n,9) = A070998(n); L(n,-9) = A057081(n)*(-1)^n; %C A108299 L(n,10) = A072256(n+1); L(n,-10) = A054320(n)*(-1)^n; %C A108299 L(n,11) = A078922(n+1); L(n,-11) = A097783(n)*(-1)^n; %C A108299 L(n,12) = A077417(n); L(n,-12) = A077416(n)*(-1)^n; %C A108299 L(n,13) = A085260(n); %C A108299 L(n,14) = A001570(n); L(n,-14) = A028230(n)*(-1)^n; %C A108299 L(n,n) = A108366(n); L(n,-n) = A108367(n). %C A108299 Row n of the matrix inverse (A124645) has g.f.: x^[n/2]*(1-x)^(n-[n/2]). - Paul D. Hanna (pauldhanna(AT)juno.com), Jun 12 2005 %D A108299 Friedrich L. Bauer, 'De Moivre und Lagrange: Cosinus eines rationalen Vielfachen von Pi', Informatik Spektrum 28 (Springer, 2005). %F A108299 T(n+1, k) = if sign(T(n, k-1))=sign(T(n, k)) then T(n, k-1)+T(n, k) else -T(n, k-1) for 0