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a(0)=a(1)=a(2)=0, a(3)=1; thereafter a(n) = 4a(n-3)-5a(n-4).
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%I #6 May 01 2013 21:16:02

%S 0,0,0,1,0,0,4,-5,0,16,-40,25,64,-240,300,131,-1280,2400,-976,-5775,

%T 16000,-15904,-18220,92875,-143616,6640,462600,-1038839,744640,

%U 1817200,-6468356,8172755,3545600,-34959424,65032800,-26681375,-157565696,434928320,-431889500,-496855909,2527541760

%N a(0)=a(1)=a(2)=0, a(3)=1; thereafter a(n) = 4a(n-3)-5a(n-4).

%D G. Mantel, Resten van wederkeerige Reeksen, Nieuw Archief v. Wiskunde, 2nd series, I (1894), 172-184. [From _N. J. A. Sloane_, Dec 17 2010]

%K sign

%O 0,7

%A _N. J. A. Sloane_, Dec 19 2010