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%I #12 Oct 13 2022 12:53:57
%S 0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,2,1,0,0,0,0,0,0,0,1,2,3,2,1,0,0,0,0,0,
%T 0,0,0,0,0,0,1,2,3,4,5,4,3,2,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,2,
%U 3,4,5,6,7,8,7,6,5,4,3,2,1,0,0,0,0,0,0
%N a(n) = b(n) - b(b(n)) - b(n - b(n)) for n >= 3, where b(n) = A356988(n).
%C Starting at n = 7, the sequence consists of successive blocks of integers of the form 1, 2, 3, ..., F(k) - 1, F(k), F(k) - 1, ..., 3, 2, 1, where F(k), k >= 1, denotes the k-th Fibonacci number, followed by a string of zeros conjecturally of length 1 + 2*F(k+1).
%C The sequence has local peak values at abscissa values n = 7, 11, 18, ..., L(k), ..., where L(k) = A000032(k), the k-th Lucas number. The zero strings begin at abscissa values n = 8, 12, 20, 32, 52, ..., equal to the sequence {L(k) + F(k-3) : k >= 4} = {4*F(k-1): k >= 4}.
%F a(n+1) - a(n) is in {1, 0, -1}.
%F For k >= 3, a(L(k) + j) = F(k-3) - j and a(L(k) - j) = F(k-3) - j for 0 <= j <= F(k-3), where F(k) = A000045(k), the k-th Fibonacci number and L(k) = A000032(k), the k-th Lucas number.
%e Sequence {a(n)} arranged as a sequence of strings of length 2*Fibonacci(k), k >= 1
%e 0, 0;
%e 0, 0;
%e 1, 0, 0, 0;
%e 1, 0, 0, 0, 0, 0;
%e 1, 2, 1, 0, 0, 0, 0, 0, 0, 0;
%e 1, 2, 3, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
%e 1, 2, 3, 4, 5, 4, 3, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
%e 1, 2, 3, 4, 5, 6, 7, 8, 7, 6, 5, 4, 3, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
%e ...
%p # b(n) = A356988(n)
%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(b(n)) - b(n - b(n)), n = 3..250);
%Y Cf. A000032, A000045, A356988, A356991 - A356999.
%K nonn,easy
%O 3,16
%A _Peter Bala_, Sep 09 2022
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