|
Search: id:A080426
|
|
|
| A080426 |
|
a(1)=1, a(2)=3; all terms are either 1 or 3; each run of 3's is followed by a run of two 1's; and a(n) is the length of the n-th run of 3's. |
|
+0
5
|
|
| 1, 3, 1, 1, 3, 3, 3, 1, 1, 3, 1, 1, 3, 1, 1, 3, 3, 3, 1, 1, 3, 3, 3, 1, 1, 3, 3, 3, 1, 1, 3, 1, 1, 3, 1, 1, 3, 3, 3, 1, 1, 3, 1, 1, 3, 1, 1, 3, 3, 3, 1, 1, 3, 1, 1, 3, 1, 1, 3, 3, 3, 1, 1, 3, 3, 3, 1, 1, 3, 3, 3, 1, 1, 3, 1, 1, 3, 1, 1, 3, 3, 3, 1, 1, 3, 3, 3, 1, 1, 3, 3, 3, 1, 1, 3, 1, 1, 3, 1, 1, 3, 3, 3, 1, 1
(list; graph; listen)
|
|
|
OFFSET
|
1,2
|
|
|
COMMENT
|
It appears that the sequence can be calculated by any of the following three methods: (1) Start with 1 and repeatedly replace (simultaneously) all 1's with 1,3,1 and all 3's with 1,3,3,3,1. (2) a(n)= A026490(2n). (3) Replace each 2 in A026465 with 3.
Length of n-th run of 1's in the Feigenbaum sequence A035263 = 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, .... - DELEHAM Philippe (kolotoko(AT)wanadoo.fr), Apr 18 2004
|
|
FORMULA
|
a(1) = 1; for n>1, a(n) = A003156(n) - A003156(n-1). - DELEHAM Philippe (kolotoko(AT)wanadoo.fr), Apr 16 2004
|
|
CROSSREFS
|
Cf. A026465, A026490.
Adjacent sequences: A080423 A080424 A080425 this_sequence A080427 A080428 A080429
Sequence in context: A094782 A035666 A060592 this_sequence A133116 A059959 A051120
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
John W. Layman (layman(AT)math.vt.edu), Feb 18 2003
|
|
|
Search completed in 0.004 seconds
|
