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A078338 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,u(k)). 0
1, 2, 3, 3, 4, 5, 5, 4, 5, 5, 4, 3, 3, 2, 1, 1, 2, 1, 1, 2, 3, 3, 4, 5, 5, 4, 5, 5, 4, 3, 3, 2, 1, 1, 2, 1, 1, 2, 3, 3, 4, 5, 5, 4, 5, 5, 4, 3, 3, 2, 1, 1, 2, 1, 1, 2, 3, 3, 4, 5, 5, 4, 5, 5, 4, 3, 3, 2, 1, 1, 2, 1, 1, 2, 3, 3, 4, 5, 5, 4, 5, 5, 4, 3, 3, 2, 1, 1, 2, 1, 1, 2, 3, 3, 4, 5, 5, 4, 5, 5, 4, 3, 3, 2, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

LINKS

Table of n, a(n) for n=1..105.

Index entries for linear recurrences with constant coefficients, signature (1, 0, 0, 0, 0, 0, 0, 0, -1, 1).

FORMULA

Periodic with period 18.

a(n)=(1/306)*{6*[n mod 18] + 23*[(n + 1) mod 18] - 11*[(n + 2) mod 18] + 6*[(n + 3) mod 18] + 23*[(n + 4) mod 18] + 23*[(n + 5) mod 18] + 6*[(n + 6) mod 18] + 23*[(n + 7) mod 18] + 23*[(n + 8) mod 18] + 6*[(n + 9) mod 18] - 11*[(n + 10) mod 18] + 23*[(n + 11) mod 18] + 6*[(n + 12) mod 18] - 11*[(n + 13) mod 18] - 11*[(n + 14) mod 18] + 6*[(n + 15) mod 18] - 11*[(n + 16) mod 18] - 11*[(n + 17) mod 18]}, with n>=0. - Paolo P. Lava, Jun 01 2007

MATHEMATICA

LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 0, -1, 1}, {1, 2, 3, 3, 4, 5, 5, 4, 5, 5}, 105] (* Ray Chandler, Aug 27 2015 *)

CROSSREFS

Sequence in context: A179840 A240153 A241435 * A278149 A007306 A286544

Adjacent sequences:  A078335 A078336 A078337 * A078339 A078340 A078341

KEYWORD

nonn

AUTHOR

Benoit Cloitre, Nov 21 2002

STATUS

approved

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Last modified November 17 00:08 EST 2019. Contains 329209 sequences. (Running on oeis4.)