Simple(?) Fib problem

[Ron]

David Wilson asked:

> For which m for do the Fibs have every residue mod m?

A quick search reveals:
1 2 3 4 5 6 7 9 10 14 15 20 25 27 30 35 45 50 70 75 81 100 125 135 150
175 225 243 250 350 375 405 500 625 675 729 750 875 1125 1215 1250 1750
1875 2025 2187 2500 3125 3375 3645 3750 4375 5625 6075 6250 6561 8750
9375 
 (extended after 500 by David Wilson)
 
 The number of different residues of Fibonacci's mod M for M from 1 to
100 is:
1,2,3,4,5,6,7,6,9,10,7,11,9,14,15,11,13,11,12,20,9,14,19,13,25,18,27,21,
10,30,19,21,19,13,35,15,29,13,25,30,19,18,33,20,45,21,15,15,37,50,35,30,
37,29,12,25,33,20,37,55,25,21,23,42,45,38,51,20,29,70,44,15,57,58,75,13,
35,50,49,55,81,38,63,27,65,66,27,32,17,55,53,29,53,30,60,26,69,74,41,100

So the i for which a[i]=i in the latter series make up the former
series.

Both of these are have now been submitted to the OEIS as A102380 and
A102381

Ron Knott

www.mcs.surrey.ac.uk/Personal/R.Knott/Fibonacci