%I #19 Jul 19 2015 10:44:50

%S 4,5,4,6,4,5,7,11,10,5,6,8,10,14,11,21,14,6,7,9,21,13,15,23,14,5,20,

%T 33,41,17,7,8,10,20,35,14,14,23,16,33,31,51,20,33,13,32,50,45,53,24,

%U 44,8,9,11,17,33,35,12,4,35,40,17,47,48,31,14,25,20,40,27

%N Hiccup sequence.

%C The hiccup sequence #_n: Consider the sequences created by a_n = a_[n-1] + a_[n-2] in Z mod n, with a certain "carrying" rule, where when the numbers a_[n-2] and a_[n-1] are added together in Z and the result is greater than n, two numbers are added to the sequence, first 1, and then a_[n-2] + a_[n-1] mod n. These sequences all fall into cycles of a different length:

%C 2: 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0... [4]

%C 3: 0, 1, 1, 2, 1, 0, 1, 1, 2, 1, 0, 1, 1, 2, 1, 0, 1... [5]

%C 4: 0, 1, 1, 2, 3, 1, 1, 2, 3, 1, 1, 2, 3, 1, 1, 2, 3... [4]

%C 5: 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3... [6]

%C 6: 0, 1, 1, 2, 3, 5, 1, 2, 3, 5, 1, 2, 3, 5, 1, 2, 3... [4]

%C 7: 0, 1, 1, 2, 3, 5, 1, 1, 2, 3, 5, 1, 1, 2, 3, 5, 1... [5]

%C 8: 0, 1, 1, 2, 3, 5, 1, 0, 1, 1, 2, 3, 5, 1, 0, 1, 1... [7]

%C ...and so on! It is easy to calculate and it seems to go on forever. But I do not have a proof.

%C Proof: This sequence continues forever. This is because the state of the sequence is totally determined by the previous two numbers, and it has only a finite number of states. Therefore, the same state of any carrying Fibonacci sequence must repeat after a finite period of time; in physics, this is Poincaré's recurrence theorem.

%C Proof: This sequence increases without bound. This is because in order to cycle a carrying Fibonacci sequence must wrap around at least once. Thus

%C #_n ≥ sup{k in N, Fib[k] < n}

%H Lars Blomberg, <a href="/A248624/b248624.txt">Table of n, a(n) for n = 2..10000</a>

%o (Lua)

%o for i = 2, 100 do

%o local a = {i}

%o for j = 1, 20 do

%o local k

%o if j <3 then

%o k = j-1

%o else

%o k = a[#a] + a[#a-1]

%o end

%o if k >= i then

%o a[#a+1] = 1

%o k = k - i

%o end

%o a[#a+1] = k

%o end

%o print(table.concat(a, ", "))

%o end

%Y Cf. A248218, A001175.

%K nonn

%O 2,1

%A _Mason Bogue_, Oct 10 2014

%E Corrected a(9) and added a(13)-a(71) by _Lars Blomberg_, Oct 18 2014