A356988 a(n) = n - a^[2](n - a^[3](n-1)) with a(1) = 1, where a^[2](n) = a(a(n)) and a^[3](n) = a(a(a(n))).

1, 1, 2, 3, 3, 4, 5, 5, 6, 7, 8, 8, 8, 9, 10, 11, 12, 13, 13, 13, 13, 14, 15, 16, 17, 18, 19, 20, 21, 21, 21, 21, 21, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 34, 34, 34, 34, 34, 34, 34, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 55, 55, 55, 55, 55

Offset: 1

Author

Peter Bala, Sep 08 2022

Keywords

• nonn
• easy

Comments

This is the second sequence in a family of nested-recurrent sequences with apparently similar structure defined as follows. Given a sequence s = {s(n) : n >= 1} we define the k-th iterated sequence s^[k] by putting s^[1](n) = s(n) and setting s^[k](n) = s^[k-1](s(n)) for k >= 2. For k >= 1, we define a nested-recurrent sequence, dependent on k, by putting u(1) = 1 and setting u(n) = n - u^[k](n - u^[k+1](n-1)) for n >= 2. This is the case k = 2. For other cases see A006165 (k = 1), A356989 (k = 3) and A356990 (k = 4).

The sequence is slow, that is, for n >= 1, a(n+1) - a(n) is either 0 or 1. The sequence is unbounded.

The line graph of the sequence {a(n)} thus consists of a series of plateaus (where the value of the ordinate a(n) remains constant as n increases) joined by lines of slope 1.

The sequence of plateau heights begins 3, 5, 8, 13, 21, 34, 55, ..., the Fibonacci numbers A000045.

The plateaus start at abscissa values n = 4, 7, 11, 18, 29, 47, 76, ..., the Lucas numbers A000032, and finish at abscissa values n = 5, 8, 13, 21, 34, 55, 89, ..., the Fibonacci numbers. The sequence of plateau lengths 1, 1, 2, 3, 5, 8, 13, ... is thus the Fibonacci sequence.

The iterated sequences{a^[k](n) : n >= 1}, k = 2, 3,..., share similar properties to the present sequence. See the Example section below.

Examples

			Related sequences:
1) The square of the sequence: {a^[2](n) : n >= 1} = {a(a(n)) : n >= 1}. The first few terms are
  1, 1, 1, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 7, 8, 8, 8, 8, 8, 8, 9, 10, 11, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 15, 16, 17, 18, 19, 20, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 35, 36, 37, 38, 39, ...
The sequence is slow. The line graph of the sequence has plateaus of height Fibonacci(k), k >= 2, starting at abscissa value 2*Fibonacci(k) and ending at abscissa Fibonacci(k+2).
2) The cube of the sequence: {a^[3](n) : n >= 1} = {a(a(a(n))) : n >= 1}. The first few terms are
  1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 4, 5, 5, 5, 5, 5, 5, 5, 6, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 10, 11, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 15, 16, 17, 18, 19, 20, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 22, 23, 24, 25, 26, 27, ...
The line graph of the sequence has plateaus of height Fibonacci(k), k >= 2, starting at abscissa value 3*Fibonacci(k) and ending at abscissa Fibonacci(k+3).
		

Links

• Peter Bala, Notes on A356988

Crossrefs

Cf. A000032, A000045, A001611, A006165, A022114, A022120, A022367, A104449, A356989, A356990, A356991 - A356999.

Programs

Maple

a := proc(n) option remember; if n = 1 then 1 else n - a(a(n - a(a(a(n-1))))) end if; end proc:
seq(a(n), n = 1..100);

Formula

a(n+1) - a(n) = 0 or 1.

The terms of the sequence are completely determined by the following two results:

a) for n >= 2, a(L(n-1) + j) = F(n) for 0 <= j <= F(n-3), where F(n) = A000045(n), the n-th Fibonacci number with F(-1) = 1 and L(n) = A000032(n), the n-th Lucas number;

b) for n >= 2, a(F(n+1) + j) = F(n) + j for 0 <= j <= F(n-1).

Hence a(F(n+2)) = a(F(n+1)) + a(F(n)) for n >= 2 and a(L(n+2)) = a(L(n+1)) + a(L(n)) for n >= 0.

a(2*F(n)) = Lucas(n-1) for n >= 2;

a(3*F(n)) = 2*F(n) for n >= 1;

a(4*F(n)) = F(n+2) for n >= 2;

a(5*F(n)) = 4*F(n) - F(n-1) = A022120(n-2) for n >= 2.

a(2*L(n)) = F(n) + 3*F(n-1) = A104449(n) for n >= 0;

a(3*L(n)) = F(n+3) for n >= 3;

a(4*L(n)) = F(n+4) - L(n-3) = A022114(n-1) for n >= 3;

a(5*L(n)) = 11*F(n-1) + F(n-4) = A022367(n-1) for n >= 4.

For n >= 1, m >= 2, a(F(m*n)) = F(m*n-1) and a(L(m*n)) = F(m*n+1). Hence

a(L(m*n)) + a(F(m*n)) = L(m*n) and a(L(m*n)) - a(F(m*n)) = F(m*n).

Conjectures:

1) a(n) + a^[2](n - a^[2](n - a^[2](n))) = n for n >= 2.

2) If k >= 2 and m = 2*k - 1 then a(m*n - a(k*n)) = a(m*n - a(m*n - a(m*n - a(k*n)))).