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%I #10 Oct 23 2022 23:35:24
%S 0,1,1,0,1,1,0,1,2,2,2,1,0,1,2,3,3,3,3,2,1,0,1,2,3,4,5,5,5,5,5,5,4,3,
%T 2,1,0,1,2,3,4,5,6,7,8,8,8,8,8,8,8,8,8,7,6,5,4,3,2,1,0,1,2,3,4,5,6,7,
%U 8,9,10,11,12,13,13,13,13,13,13,13,13,13,13,13,13,13,13,12,11,10,9,8,7,6,5,4,3,2
%N a(n) = b(n) - 2*b(b(b(n))) for n >= 3, where b(n) = A356988(n).
%C a(n+1) - a(n) is equal to 0, 1 or -1.
%C The sequence vanishes at abscissa values n = 3, 6, 9, 15, 24, 39, ..., 3*Fibonacci(k), ....
%C For k >= 2, the line graph of the sequence, starting from the zero value at abscissa n = 3*Fibonacci(k), ascends with slope 1 to a local plateau at height Fibonacci(k-1) at abscissa value n = Lucas(k+1). The plateau has length Fibonacci(k-1). From the end of the plateau, at abscissa value n = Fibonacci(k+3), the graph of the sequence descends with slope -1 to the next zero at abscissa n = 3*Fibonacci(k+1).
%H Peter Bala, <a href="/A357563/a357563.pdf">Notes on A357563</a>
%F For k >= 2 there holds
%F a(3*Fibonacci(k) + j) = j for 0 <= j <= Fibonacci(k-1) (rise from 0 to plateau)
%F a(Lucas(k+1) + j) = Fibonacci(k-1) for 0 <= j <= Fibonacci(k-1) (plateau)
%F a(Fibonacci(k+3) + j) = Fibonacci(k-1) - j for 0 <= j <= Fibonacci(k-1) (descent back to 0).
%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) - 2*b(b(b(n))), n = 3..100);
%Y Cf. A000032, A000045, A356988, A357562.
%K nonn,easy
%O 3,9
%A _Peter Bala_, Oct 14 2022
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