%I #16 Nov 04 2014 22:40:35
%S 0,2,1,5,7,3,17,29,8,58,13,109,21,203,34,370,55,667,89,1189,144,2102,
%T 233,3689,377,6435,610,11166,987,19287,1597,33181,2584,56882,4181,
%U 97205,6765,165643,10946,281546,17711,477443,28657,807941,46368,1364590,75025,2300689,121393
%N Sequence of distinct least nonnegative numbers such that the average of the first n terms is a Fibonacci number.
%C This is the same concept as A248982, but with a(1) = 0, the zeroth Fibonacci number.
%F Conjecture: a(n) = 2*a(n-2) + a(n-4) - 2*a(n-6) - a(n-8) for n > 15. - _Colin Barker_, Oct 24 2014
%F Empirical g.f.: x^2*(x -1)*(12*x^12 + 28*x^10 + 3*x^9 - 2*x^8 + 7*x^7 - 24*x^6 - 2*x^5 - 9*x^3 - 4*x^2 - 3*x - 2) / (x^4 + x^2 - 1)^2. - _Colin Barker_, Oct 24 2014
%F For n > 5, a(2*n-3) = A000045(n).
%e a(4) = 5 because neither 3 nor 4 works (the average of {0, 2, 1, 3} is 3/2, while the average of {0, 2, 1, 4} is 7/4) and 5 does work, as the average of {0, 2, 1, 5} is 2, which is a Fibonacci number.
%e a(5) = 7 because neither 3, 4 nor 6 works (the averages would be 11/5, 12/5, 14/5, respectively) and 7 does work, as the average of {0, 2, 1, 5, 7} is 3, which is a Fibonacci number.
%o (PARI) v=[]; n=0; while(n<10^7, num=(vecsum(v)+n); if(num%(#v+1)==0&&vecsearch(vecsort(v), n)==0, for(i=0, n+4, if(fibonacci(i)>(num/(#v+1)), break); if(fibonacci(i)==(num/(#v+1)), print1(n, ", "); v=concat(v, n); n=0; break))); n++)
%Y Cf. A248982, A000045.
%K nonn
%O 1,2
%A _Derek Orr_, Oct 23 2014