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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

%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.

%F a(n)=(1/306)*{45*[n mod 18] + 45*[(n + 1) mod 18] + 28*[(n + 2) mod 18] + 45*[(n + 3) mod 18] + 45*[(n + 4) mod 18] + 28*[(n + 5) mod 18] + 11*[(n + 6) mod 18] + 11*[(n + 7) mod 18] - 6*[(n + 8) mod 18] - 23*[(n + 9) mod 18] - 23*[(n + 10) mod 18] - 6*[(n + 11) mod 18] - 23*[(n + 12) mod 18] - 23*[(n + 13) mod 18] - 6*[(n + 14) mod 18] + 11*[(n + 15) mod 18] + 11*[(n + 16) mod 18] + 28*[(n + 17) mod 18]}, with n>=0. - _Paolo P. Lava_, Jun 08 2007

%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 September 22 15:04 EDT 2021. Contains 347607 sequences. (Running on oeis4.)