|
Search: id:A078339
|
|
|
| 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 1
|
|
| 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, 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, 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
(list; graph; listen)
|
|
|
OFFSET
|
1,6
|
|
|
COMMENT
|
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.
|
|
FORMULA
|
Periodic with period 18.
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 (ppl(AT)spl.at), Jun 08 2007
|
|
CROSSREFS
|
Sequence in context: A099441 A129359 A110801 this_sequence A047220 A064994 A138235
Adjacent sequences: A078336 A078337 A078338 this_sequence A078340 A078341 A078342
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Benoit Cloitre (benoit7848c(AT)orange.fr), Nov 21 2002
|
|
|
Search completed in 0.002 seconds
|