0,3
For n>1, a(n) = Fibonacci(n-1) + Lucas(n) - [3+(-1)^n]/2. - R. Stephan, May 13 2004
Sequence in context: A165753 A090752 A051058 this_sequence A026691 A018150 A019471
Adjacent sequences: A026622 A026623 A026624 this_sequence A026626 A026627 A026628
nonn
Clark Kimberling (ck6(AT)evansville.edu)
Search completed in 0.002 seconds