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%I #19 Dec 10 2023 09:24:44
%S 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 11,11,11,11,12,13,13,13,13,13,13,13,13,13,14,15,16,17,18,18,18,18,18,
%U 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 a(n) = b(n - b(n - b(n - b(n)))) for n >= 2, where b(n) = A356988(n).
%C The sequence is slow, that is, for n >= 2, a(n+1) - a(n) is either 0 or 1. The sequence is unbounded.
%C 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 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 Peter Bala, <a href="/A356993/a356993.pdf">Notes on A356993</a>
%F a(2) = a(3) = a(4) = a(5) = 1 and then for k >= 3 there holds
%F a(3*F(k) + j) = F(k) for 0 <= j <= F(k-1) (local plateau)
%F a(L(k+1) + j) = F(k) + j for 0 <= j <= F(k-2) (ascent to plateau of height L(k-1))
%F a(4*F(k) + j) = L(k-1) for 0 <= j <= F(k-1) (local plateau)
%F 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 # b(n) = A356988
%p 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 seq( b(n - b(n - b(n - b(n)))), n = 2..100 );
%Y Cf. A000032, A000045, A000285, A022087, A356988, A356991 - A356999.
%K nonn,easy
%O 2,5
%A _Peter Bala_, Sep 09 2022
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