cp's OEIS Frontend

A356994 a(n) = n - b(b(b(n))), where b(n) = A356988(n).

%I A356994 #13 Aug 10 2023 01:54:36
%S A356994 0,1,2,3,4,4,5,6,6,7,8,9,10,10,10,11,12,13,14,15,16,16,16,16,17,18,19,
%T A356994 20,21,22,23,24,25,26,26,26,26,26,26,27,28,29,30,31,32,33,34,35,36,37,
%U A356994 38,39,40,41,42,42,42,42,42,42,42,42,42,43,44,45
%N A356994 a(n) = n - b(b(b(n))), where b(n) = A356988(n).
%C A356994 The sequence is slow, that is, for n >= 2, a(n+1) - a(n) is either 0 or 1. The sequence is unbounded.
%C A356994 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.
%C A356994 The sequence of plateau heights begins 2, 4, 6, 10, 16, 26, 42, 68, 110, ..., the sequence {2*Fibonacci(k): k >= 2}
%C A356994 The plateau of height 2*F(k), k >= 2, has length equal to Fibonacci(k-2), starting at abscissa value n = Fibonacci(k+2) and ending at abscissa n = 3*Fibonacci(k).
%F A356994 The sequence is determined by the initial values a(1) = 0, a(2) = 1 and the pair of formulas
%F A356994 1) a(n) = 2*Fibonacci(k) for n in the integer interval [Fibonacci(k+2), 3*Fibonacci(k)], k >= 2, and
%F A356994 2) for k >= 2, a(3*Fibonacci(k) + j) = 2*Fibonacci(k) + j for 0 <= j <= 2*Fibonacci(k-1).
%p A356994 #  b(n) = A356988
%p A356994 b := proc(n) option remember; if n = 1 then 1 else n - b(b(n - b(b(b(n-1))))) end if; end proc:
%p A356994 seq( n - b(b(b(n))), n = 1..100);
%Y A356994 Cf. A000045, A356988, A356991, A356992, A356993, A356995, A356996, A356997, A356998, A356999.
%K A356994 nonn,easy
%O A356994 1,3
%A A356994 _Peter Bala_, Sep 09 2022