Mason Bogue - cp's OEIS frontend

User: Mason Bogue

Mason Bogue has authored 1 sequences.

A248624 Hiccup sequence.

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, 33, 41, 17, 7, 8, 10, 20, 35, 14, 14, 23, 16, 33, 31, 51, 20, 33, 13, 32, 50, 45, 53, 24, 44, 8, 9, 11, 17, 33, 35, 12, 4, 35, 40, 17, 47, 48, 31, 14, 25, 20, 40, 27

Offset: 2

Mason Bogue, Oct 10 2014

nonn

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:
2: 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0... [4]
3: 0, 1, 1, 2, 1, 0, 1, 1, 2, 1, 0, 1, 1, 2, 1, 0, 1... [5]
4: 0, 1, 1, 2, 3, 1, 1, 2, 3, 1, 1, 2, 3, 1, 1, 2, 3... [4]
5: 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3... [6]
6: 0, 1, 1, 2, 3, 5, 1, 2, 3, 5, 1, 2, 3, 5, 1, 2, 3... [4]
7: 0, 1, 1, 2, 3, 5, 1, 1, 2, 3, 5, 1, 1, 2, 3, 5, 1... [5]
8: 0, 1, 1, 2, 3, 5, 1, 0, 1, 1, 2, 3, 5, 1, 0, 1, 1... [7]
...and so on! It is easy to calculate and it seems to go on forever. But I do not have a proof.
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.
Proof: This sequence increases without bound. This is because in order to cycle a carrying Fibonacci sequence must wrap around at least once. Thus
  #_n ≥ sup{k in N, Fib[k] < n}

Lars Blomberg, Table of n, a(n) for n = 2..10000

Cf. A248218, A001175.

for i = 2, 100 do
  local a = {i}
  for j = 1, 20 do
      local k
      if j <3 then
         k = j-1
      else
         k = a[#a] + a[#a-1]
      end
      if k >= i then
         a[#a+1] = 1
         k = k - i
      end
      a[#a+1] = k
  end
    print(table.concat(a, ", "))
end

Corrected a(9) and added a(13)-a(71) by Lars Blomberg, Oct 18 2014

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User: Mason Bogue

A248624 Hiccup sequence.

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