Search: id:A080426 Results 1-1 of 1 results found. %I A080426 %S A080426 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, %T A080426 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, %U A080426 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 %N 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. %C A080426 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. %C A080426 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 %F A080426 a(1) = 1; for n>1, a(n) = A003156(n) - A003156(n-1). - DELEHAM Philippe (kolotoko(AT)wanadoo.fr), Apr 16 2004 %Y A080426 Cf. A026465, A026490. %Y A080426 Sequence in context: A094782 A035666 A060592 this_sequence A133116 A059959 A051120 %Y A080426 Adjacent sequences: A080423 A080424 A080425 this_sequence A080427 A080428 A080429 %K A080426 nonn %O A080426 1,2 %A A080426 John W. Layman (layman(AT)math.vt.edu), Feb 18 2003 Search completed in 0.001 seconds