Concerning A112237.

[koh]

    Hi,Bob
    Thanks for trying to extend my sequence.
    I think the definition of A112237 became better.
    So, I suppose you will understand it.

    I explain the reason why I came across these generalizations of Fibonacci.
    We know two types of numbers exist.
    One is ordinal number and another is cardinal number.
    So, the definition of Fibonacci number means the following four cases.

         F(n)=F(n-2)+F(n-1) 

        1.      F(n-2) is cardinal, F(n-1) is cardinal
        2.      F(n-2) is cardinal, F(n-1) is ordinal
        3.      F(n-2) is ordinal, F(n-1) is cardinal
        4.      F(n-2) is ordinal, F(n-1) is ordinal

    If we calculate these number on the sequence of natural number, then the results are the same, but if we calculate them on the other sequences, then the results are different.

        1.      F(n)=F(n-2)+F(n-1)
        2.      a(n)=Fibonacci(#Fibonacci(a(n-2))+a(n-1))
        3.      a(n)=Fibonacci(a(n-2)+#Fibonacci(a(n-1))) 
        4.      a(n)=Fibonacci(#Fibonacci(a(n-2))+#Fibonacci(a(n-1))) 
                     =Fibonacci(Fibonacci(n)) 
                Where #Fibonacci(n) means number of Fibonacci numbers <=n.
                If n is i-th Fibonacci number then #Fibonacci(n) gives i.

    The 2nd case is A112237.
      .
     Fibonacci : 1,2,3,5,8,13,21,34,55,89,144
        a(n)    : 1,2,3,8,144 

        a(5)=Fibonacci(#Fibonacci(a(5-2))+a(5-1))
             =Fibonacci(#Fibonacci(3)+8)
             =Fibonacci(3+8)
             =144
             where Fibonacci(1)=1, Fibonacci(2)=2, which are different from the standard condition that Fibonacci(1)=0, Fibonacci(2)=1 .       
             condition 2 says that 8 is ordinal number, so we get 8th term from third term.

    The 4th case is not so interesting.

    The 3rd case is the following.

         %I A000001
         %S A000001 1,2,3,8,34,
         %N A000001 If a(n-1) is the i-th Fibnacci number then a(n)=Fibonacci(i+a(n-2)); with a(1)=1, a(2)=2, and where we use the following nonstandard indexing for the Fibonacci numbers: f(n)=f(n-1)+f(n-2), f(1)=1, f(2)=2 (cf. A000045).
         %e A000001 a(5)=Fibonacci(5+3)=34 because a(4) is 5th Fibonacci number and a(3)=3 
         %Y A000001 A112237, A000045, A112601.
         %K A000001 nonn
         %O A000001 1,2
         %A A000001 Yasutsohi Kohmoto zbi74583(AT)boat.zero.ad.jp

    Yasutoshi
    

>Dear Sir,
>
>	I am trying to extend this sequence and am having a terrible time doing so.
>Can you please help?
>
>	a(5)=Fibonacci(InverseFibonacci(a(3))+a(4))=Fibonacci(InverseFibonacci(3+8)=
>=Fibonacci(InverseFibonacci(11)=Fibonacci(89)=1779979416004714189 and not 144.
>
>	Where am I going wrong!
>
>Sincerely, Bob.
>
>
>%I A112237
>%S A112237 1,2,3,8,144
>%N A112237 a(n)=Fibonacci(InverseFibonacci(a(n-2))+a(n-1)).
>%Y A112237 Cf. A000045.
>%K A112237 nonn,uned,obsc,more,new
>%O A112237 1,2
>%A A112237 Yasutsohi Kohmoto zbi74583(AT)boat.zero.ad.jp, Dec 22 2005
>