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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

Table of n, a(n) for n=1..105.

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 August 6 12:33 EDT 2020. Contains 336246 sequences. (Running on oeis4.)