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A078339 Let u(1)=u(2)=u(3)=1 and u(n)=(-1)^n*sign(u(n-1)-u(n-2))*u(n-3); then a(n)=sum(k=1,n,sum(i=1,k,u(i))) - 3*(n-1). 0

%I #15 Jan 01 2024 13:42:27

%S 1,0,0,0,1,3,5,6,8,10,11,11,11,10,8,6,5,3,1,0,0,0,1,3,5,6,8,10,11,11,

%T 11,10,8,6,5,3,1,0,0,0,1,3,5,6,8,10,11,11,11,10,8,6,5,3,1,0,0,0,1,3,5,

%U 6,8,10,11,11,11,10,8,6,5,3,1,0,0,0,1,3,5,6,8,10,11,11,11,10,8,6,5,3,1

%N Let u(1)=u(2)=u(3)=1 and u(n)=(-1)^n*sign(u(n-1)-u(n-2))*u(n-3); then a(n)=sum(k=1,n,sum(i=1,k,u(i))) - 3*(n-1).

%C Note the palindromic form of the periodic part: 0,0,1,3,5,6,8,10,11,11,11,10,8,6,5,3,1,0,0.

%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (1, 0, 0, 0, 0, 0, 0, 0, -1, 1).

%F Periodic with period 18.

%t LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 0, -1, 1},{1, 0, 0, 0, 1, 3, 5, 6, 8, 10},91] (* _Ray Chandler_, Aug 27 2015 *)

%K nonn

%O 1,6

%A _Benoit Cloitre_, Nov 21 2002

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Last modified April 25 01:06 EDT 2024. Contains 371964 sequences. (Running on oeis4.)