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 A140345 a(n)=a(n-1)^2-a(n-2)-a(n-3)-a(n-4), a(1)=a(2)=a(3)=a(4)=1. 0
 1, 1, 1, 1, -2, 1, 1, 1, 1, -2, 1, 1, 1, 1, -2, 1, 1, 1, 1, -2, 1, 1, 1, 1, -2, 1, 1, 1, 1, -2, 1, 1, 1, 1, -2, 1, 1, 1, 1, -2, 1, 1, 1, 1, -2, 1, 1, 1, 1, -2, 1, 1, 1, 1, -2, 1, 1, 1, 1, -2, 1, 1, 1, 1, -2, 1, 1, 1, 1, -2, 1, 1, 1, 1, -2, 1, 1, 1, 1, -2, 1, 1, 1, 1, -2, 1, 1, 1, 1, -2, 1, 1, 1, 1, -2, 1, 1, 1, 1, -2, 1, 1, 1, 1, -2 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,5 COMMENTS Periodic with period 5: 1,1,1,1,-2 LINKS Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 1). FORMULA a(n)=a(n-1)^2-a(n-2)-a(n-3)-a(n-4), a(1)=a(2)=a(3)=a(4)=1 a(n)=(1/25)*{16*(n mod 5)+[(n+1) mod 5]+[(n+2) mod 5]+[(n+3) mod 5]-14*[(n+4) mod 5]}, with n>=1. - Paolo P. Lava, Jun 03 2008 a(n)=3*[(n^4 mod 5)-1]+1 [From Paolo P. Lava, May 05 2010] EXAMPLE a(5)=1^2-1-1-1=-2 MATHEMATICA a[n_]:=a[n]=a[n-1]^2-a[n-2]-a[n-3]-a[a-4]; a[1]=a[2]=a[3]=a[4]=1 LinearRecurrence[{0, 0, 0, 0, 1}, {1, 1, 1, 1, -2}, 105] (* Ray Chandler, Aug 25 2015 *) CROSSREFS Sequence in context: A139549 A216915 A280569 * A177706 A130782 A055457 Adjacent sequences:  A140342 A140343 A140344 * A140346 A140347 A140348 KEYWORD sign AUTHOR Ben Branman, May 29 2008 EXTENSIONS Extended by Ray Chandler, Aug 25 2015 STATUS approved

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Last modified July 24 16:24 EDT 2021. Contains 346273 sequences. (Running on oeis4.)