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A078109 Let u(1)=u(2)=1, u(3)=2n, u(k) = abs(u(k-1)-u(k-2)-u(k-3)) and M(k)= Max( u(i) : 1<=i<=k), then for any k>=a(n), M(k)=sqrtint(k + A078108(n)) where sqrtint(x) denotes floor(sqrt(x)). 1

%I #9 Apr 08 2015 18:02:29

%S 3,10,38,10,35,66,19,150,90,30,243,159,138,270,19,186,35,178,358,127,

%T 46,334,1,23,370,438,343,182,430,46,454,470,534,30,618,734,903,570,

%U 302,571,638,30,166,822

%N Let u(1)=u(2)=1, u(3)=2n, u(k) = abs(u(k-1)-u(k-2)-u(k-3)) and M(k)= Max( u(i) : 1<=i<=k), then for any k>=a(n), M(k)=sqrtint(k + A078108(n)) where sqrtint(x) denotes floor(sqrt(x)).

%C Conjecture : a(n) always exists, a(n)/n^2 is bounded. If initial conditions are u(1)=u(2)=1, u(3)=2n+1, then u(k) reaches a 2-cycle for any k>m large enough (cf. A078098)

%Y Cf. A000196, A078108, A077623.

%K nonn

%O 1,1

%A _Benoit Cloitre_, Dec 05 2002

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