|
Search: id:A096055
|
|
|
| A096055 |
|
Let {s(i)}, i=0,1,2,... be a sequence of finite sequences with terms s(i)(j), j=1,2,3,... Start with s(0)={1}. Then, for k>0, let s(k)=s(k-1)Us(k-1) if s(k-1)(k)=0, s(k)=s(k-1)U{0}Us(k-1) if s(k-1)(k)=1, where s(i)(j) is the j-th element of s(i) and U denotes concatenation of the terms of the two operands. {a(n)} is the limit of s(k) as k goes to infinity. |
|
+0
3
|
|
| 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0
(list; graph; listen)
|
|
|
OFFSET
|
1,1
|
|
|
COMMENT
|
Suggested by Leroy Quet Jul 18,2004.
|
|
LINKS
|
Leroy Quet, Home Page (listed in lieu of email address)
|
|
EXAMPLE
|
Let s(0)={1}. Then
s(1)=s(0)U{0}Us(0)={1,0,1}, since s(0)(1)=1,
s(2)=s(2)Us(2)={1,0,1,1,0,1}, since s(1)(2)=0,
s(3)=s(2)U{0}Us(2)={1,0,1,1,0,1,0,1,0,1,1,0,1}, since s(2)(3)=1, etc.
|
|
CROSSREFS
|
Sequence in context: A104974 A024711 A128174 this_sequence A125144 A115198 A005614
Adjacent sequences: A096052 A096053 A096054 this_sequence A096056 A096057 A096058
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
John W. Layman (layman(AT)math.vt.edu), Jul 20 2004
|
|
|
Search completed in 0.002 seconds
|
