[seqfan] Re: Smallest index of Fibonacci-like sequence containing n

[Charles Greathouse]

My script

a(n)=if(n<2,return(n));for(k=1,n-1,for(a=0,k-1,my(A=a,B=k-A);while(B<n,[A,B]=[B,A+B]);if(B==n,return(k))));n

agrees with your hand-computed terms, and verifies that the sequence is not
in the encyclopedia.

Charles Greathouse
Analyst/Programmer
Case Western Reserve University

On Wed, Nov 5, 2014 at 3:45 PM, Charles Greathouse <
charles.greathouse at case.edu> wrote:

> I think this sequence is interesting. Some quick observations: 0 is the
> only number with index 0, all positive integers have index at least 1. If a
> number is k times a Fibonacci number, then its index is at most k via (A,
> B) = (0, k); in particular, since 1 is a Fibonacci number, a(n) <= n. (This
> suggests A054495 as a cross-reference.)
>
> Charles Greathouse
> Analyst/Programmer
> Case Western Reserve University
>
> On Wed, Nov 5, 2014 at 1:50 PM, Allan Wechsler <acwacw at gmail.com> wrote:
>
>> Any two non-negative integers can seed a Fibonacci-like sequence, F[0] =
>> A,
>> F[1] = B, F[i+2] = F[i+1] + F[i].
>>
>> Let A+B be called the "index" of this sequence.
>>
>> Of all Fibonacci-like sequences containing, say, 18, the one with the
>> smallest index is {2,1,3,4,7,11,18...}, with an index of 3. So I say A[18]
>> = 3.
>>
>> If n is a classic Fibonacci number, A[n] = 1. If n is a Lucas number (like
>> 18), then A[n] = 3. If n is twice a Fibonacci number (like 16) then A[n] =
>> 2.
>>
>> I have calculated A[n] by hand for n from 1 to 24. It is quite possible
>> that I have made mistakes, but the sequence I get is:
>>
>> {1,1,1,2,1,2,3,1,3,2,3,4,1,4,3,2,5,3,5,4,1,6,4,3, ...}
>>
>> This is not in OEIS. I would've submitted it, but I would like somebody
>> else to check my arithmetic first, because it seems unlikely that such a
>> simple concept wouldn't have been entered already. If nobody steps up to
>> the place quickly I will cobble together some code and submit anyway.
>> Thanks for your assistance, seqfans!
>>
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>
>