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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)). 6

%I #10 Sep 01 2020 12:12:48

%S 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,

%T 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,

%U 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

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

%C 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

%H Reinhard Zumkeller, <a href="/A077653/b077653.txt">Table of n, a(n) for n = 1..10000</a>

%F 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

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

%o (Haskell)

%o a077653 n = a077653_list !! (n-1)

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

%o a077653_list (tail a077653_list) (drop 2 a077653_list)

%o -- _Reinhard Zumkeller_, Oct 11 2014

%Y Cf. A077623, A079623, A079624, A080096, A088226.

%K nonn

%O 1,2

%A _Benoit Cloitre_, Dec 02 2002

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Last modified April 25 01:35 EDT 2024. Contains 371964 sequences. (Running on oeis4.)