A356991 a(n) = b(n) + b(n - b(n)) for n >= 2, where b(n) = A356998(n).

2, 3, 4, 4, 5, 6, 7, 8, 9, 10, 11, 11, 12, 13, 14, 15, 16, 17, 18, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 29, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 47, 47, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 76, 76, 76, 76, 76, 77, 78, 79, 80

Offset: 2

Author

Peter Bala, Sep 08 2022

Keywords

• nonn
• easy

Comments

The sequence is slow, that is, for n >= 2, 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) is unchanged with increasing values of the abscissa n) joined by lines of slope 1.

The sequence of plateau heights begins 4, 7, 11, 18, 29, 47, 76, 123, 199, ..., the Lucas sequence {A000032(k): k >= 3}. The plateaus start at absiccsa values n = 4, 8, 12, 20, 32, 52, 84, 136, ..., the sequence {A022087(k): k >= 2}, and end at abscissa values n = 5, 8, 13, 21, 34, 55, 89, ..., the Fibonacci sequence {A000045(k): k >= 5}.

Compare with A356992 and A356993.

Other sequences defined in terms of b(n) = A356998(n) that are similarly related to the Lucas numbers include {n - b(b(b(2*n - b(n)))): n >= 1} beginning 0, 1, 2, 3, 3, 4, 4, 5, 6, 7, 7, 7, 8, 9, 10, 11, 11, 11, 11, 12, 13, 14, 15, 16, 17, 18, 18, 18, 18, 17, 18, 19, ... and {2*n - b(2*n - b(2*n - b(n))) : n >= 1} beginning 1, 3, 4, 5, 7, 7, 9, 11, 11, 12, 14, 16, 18, 18, 18, 19, 21, 23, 25, 27, 29, 29, 29, 29, 29, 31, .... Neither sequence is slow.

Crossrefs

Cf. A000032, A000045, A022087, A356988, A356992 - A356999.

Programs

Maple

b := proc(n) option remember; if n = 1 then 1 else n - b(b(n - b(b(b(n-1))))) end if; end proc:
seq( b(n) + b(n - b(n) ), n = 2..100);

Formula

The sequence is completely determined by the initial values a(2) = 2, a(3) = 3 and the pair of formulas:

1) for k >= 3, a(4*F(k-1) + j) = L(k) for 0 <= j <= F(k-4), where F(-1) = 1 and

2) for k >= 3, a(F(k+2) + j) = L(k) + j for 0 <= j <= L(k-1).