OFFSET
0,3
COMMENTS
Conjectures from Chai Wah Wu, Jun 12 2016: (Start)
All odd numbers appear in sequence.
2n+1 appears 2n+2 times.
a((n+3)^2-8) = 2n+1 is the last time 2n+1 appears.
For n > 0, a((n-1)^2+2) = 2n+1 is the first time 2n+1 appears. (End)
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..10000
MATHEMATICA
a[0] = a[1] = 1; a[n_] := a[n] = a[n - a[n - 2]] + 2; Table[a[n], {n, 0, 200}]
CROSSREFS
KEYWORD
nonn
AUTHOR
José María Grau Ribas, Apr 04 2010
STATUS
approved