cp's OEIS Frontend

A356993 a(n) = b(n - b(n - b(n - b(n)))) for n >= 2, where b(n) = A356988(n).

%I A356993 #19 Dec 10 2023 09:24:44
%S A356993 1,1,1,1,2,2,3,3,3,3,4,4,4,5,5,5,5,6,7,7,7,7,8,8,8,8,8,8,9,10,11,11,
%T A356993 11,11,11,11,12,13,13,13,13,13,13,13,13,13,14,15,16,17,18,18,18,18,18,
%U A356993 18,18,18,18,19,20,21,21,21,21,21,21,21,21,21,21,21,21,21,21,22,23,24,25,26,27,28,29,29,29,29
%N A356993 a(n) = b(n - b(n - b(n - b(n)))) for n >= 2, where b(n) = A356988(n).
%C A356993 The sequence is slow, that is, for n >= 2, a(n+1) - a(n) is either 0 or 1. The sequence is unbounded.
%C A356993 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 A356993 The sequence of plateau heights beginning 3, 4, 5, 7, 8, 11, 13, 18, 21, 29, 34, 47, 55, ..., consists of alternating Fibonacci numbers A000045 and Lucas numbers A000032.
%H A356993 Peter Bala, <a href="/A356993/a356993.pdf">Notes on A356993</a>
%F A356993 a(2) = a(3) = a(4) = a(5) = 1 and then for k >= 3 there holds
%F A356993 a(3*F(k) + j) = F(k) for 0 <= j <= F(k-1) (local plateau)
%F A356993 a(L(k+1) + j) = F(k) + j for 0 <= j <= F(k-2) (ascent to plateau of height L(k-1))
%F A356993 a(4*F(k) + j) = L(k-1) for 0 <= j <= F(k-1) (local plateau)
%F A356993 a(4*F(k) + F(k-1) + j) = L(k-1) + j for 0 <= j <= F(k-3) (ascent to next plateau of height F(k+1)).
%p A356993 # b(n) = A356988
%p A356993 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 A356993 seq( b(n - b(n - b(n - b(n)))), n = 2..100 );
%Y A356993 Cf. A000032, A000045, A000285, A022087, A356988, A356991 - A356999.
%K A356993 nonn,easy
%O A356993 2,5
%A A356993 _Peter Bala_, Sep 09 2022