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A077653

a(1)=1, a(2)=2, a(3)=2, a(n) = abs(a(n-1) - a(n-2) - a(n-3)).

1, 2, 2, 1, 3, 0, 4, 1, 3, 2, 2, 3, 1, 4, 0, 5, 1, 4, 2, 3, 3, 2, 4, 1, 5, 0, 6, 1, 5, 2, 4, 3, 3, 4, 2, 5, 1, 6, 0, 7, 1, 6, 2, 5, 3, 4, 4, 3, 5, 2, 6, 1, 7, 0, 8, 1, 7, 2, 6, 3, 5, 4, 4, 5, 3, 6, 2, 7, 1, 8, 0, 9, 1, 8, 2, 7, 3, 6, 4, 5, 5, 4, 6, 3, 7, 2, 8, 1, 9, 0, 10, 1, 9, 2, 8, 3, 7, 4, 6, 5, 5, 6, 4, 7

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OFFSET

1,2

COMMENTS

Conjecture : let z(1)=x; z(2)=y; z(3)= z; z(n)=abs(z(n-1)-z(n-2)-z(n-3)) if z(n) is unbounded (i.e. x,y,z are such that z(n) doesn't reach a cycle of length 2), then there are 2 integers n(x,y,z) and w(x,y,z) such that M(n) = floor(sqrt(n+w(x,y,z))) for n>n(,x,y,z) where M(n) = Max ( a(k) : 1<=k<=n ). As example : w(1,2,2)=9 n(1,2,2)=4; w(1,2,4)=29 n(1,2,4)=4; w(1,2,8)=157 n(1,2,8)=9

LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

FORMULA

a(n)/sqrt(n) is bounded. More precisely, let M(n) = Max ( a(k) : 1<=k<=n ); then M(n)= floor(sqrt(n+9)) for n>4

MATHEMATICA

nxt[{a_, b_, c_}]:={b, c, Abs[c-b-a]}; NestList[nxt, {1, 2, 2}, 110][[All, 1]] (* Harvey P. Dale, Sep 01 2020 *)

PROG

(Haskell)

a077653 n = a077653_list !! (n-1)

a077653_list = 1 : 2 : 2 : zipWith3 (\u v w -> abs (w - v - u))

a077653_list (tail a077653_list) (drop 2 a077653_list)

-- Reinhard Zumkeller, Oct 11 2014