Fibonacci modulo Fibonacci

[robert44444uk]

Just playing around with Fibonaccis and I notice, at least for small Fibonaccis, that there are integers 6,9,14,15,16,17,19 which never appear as mod values for Fibonaccis mod (any Fibonacci).

For example: F(11)=89 is mod 1,2,4,1,11,5,21,34,0,89,89..... the first 13 Fibonacci numbers, 89 repeating thereafter.

Looking at it the other way around, the first 44 Fibonaccis 1,1,2,3,5,8,13,21,34,55,89,144,233.... are 1,2,3,4,5,8,13,21,34,55,0,55,55,21,76,8,84,3,87,1,88,0,88,88,87,86,84,81,76,68,55,34,0,34,34,68,13,81,5,86,2,88,1,0 mod 89 and then the pattern repeats for the next 44 Fibonaccis.

Interestingly, the series 6,9,14,15,16,17,19.. does not appear in OEIS and therefore it is unclear to me if this property has been investigated.

Anyway, I would proposed based on extremely limited observations that such integers exist and further, there are infinite integers that are never mod values.

Maybe some mathematically minded person can (i) point me to the name of this property, and its proof or disproof, or (ii) prove or disprove either of these propositions.

[R.D. Silverman]

Originally Posted by robert44444uk

Anyway, I would proposed based on extremely limited observations that such integers exist and further, there are infinite integers that are never mod values.

There is no such thing as an "infinite" integer. Perhaps you mean "infinitely
many"???

The question you ask (rephrased) is whether the range of F_n mod F_m is Z.

This is a somewhat interesting question. I will look into it if I can find
the time.

A secondary question would be to ask for the DENSITY (in Z) of the set F_n mod F_m if it is not all of Z.

[fivemack]

Originally Posted by robert44444uk

Just playing around with Fibonaccis and I notice, at least for small Fibonaccis, that there are integers 6,9,14,15,16,17,19 which never appear as mod values for Fibonaccis mod (any Fibonacci).

Well, the Fibonacci numbers have very small period modulo the Fibonacci numbers; F(2n), F(3n), F(4n) are divisible by F(n), F(4n+1) is always 1 mod F(n) so the period is at most 4n, and if n is even then F(2n+1)=1 mod F(n) so the period is 2n.

eg: mod F{14} we have

0 1 1 2 3 5 8 13 21 34 55 89 144 233
0 233 -144 89 -55 34 -21 13 -8 5 -3 2 -1 1
and then the sequence repeats

mod F{11} we have
0 1 1 2 3 5 8 13 21 34 55
0 55 -34 21 -13 8 -5 3 -2 1 -1
0 -1 -1 -2 -3 -5 -8 -13 -21 -34 -55
0 -55 34 -21 13 -8 5 -3 2 -1 1
and then the sequence repeats

So the sequence is a subset of 'what numbers are the difference of two Fibonacci numbers'; since the Fibonacci numbers grow exponentially, the density of their differences is 0. And you'll never find excitingly small differences because, if you want the difference of two Fibonacci numbers to be 76 then you know that the larger can be no more than the Fibonacci number three beyond 76 (ie 233) because F_n - F_{n-m} is at least F_{n-2}. And indeed 76 = 89-13.

[robert44444uk]

Brilliant, than you Bob and Fivemack.