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%I #13 Oct 13 2022 05:55:34
%S 0,1,2,2,3,4,3,4,5,6,5,5,6,7,8,9,10,9,8,8,9,10,11,12,13,14,15,16,15,
%T 14,13,13,13,14,15,16,17,18,19,20,21,22,23,24,25,26,25,24,23,22,21,21,
%U 21,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,41,40,39,38,37,36,35,34,34,34,34,34,34,35,36,37
%N a(n) = b(n) - b(n - b(n)) for n >= 2, where b(n) = A356988(n).
%C The line graph of the sequence rises with slope 1 to a local peak value at heights 4, 6, 10, 16, 26, 42, ..., the sequence {2*Fibonacci(k): k >= 3}, before descending with slope -1 to a local trough at heights 3, 5, 8, 13, 21, ..., the sequence {Fibonacci(k): k >= 4}.
%C The local peaks of the graph occur at abscissa values n = 7, 11, 18, 29, 47, 76, ..., the sequence {Lucas(k): k >= 4}.
%C The trough of height F(k) starts at abscissa n = 4*F(k-1) and ends at abscissa n = F(k+2).
%C The sequence of trough lengths starting at abscissa n = 8 begin 0, 1, 1, 2, 3, 5, 8, 13, ..., the Fibonacci sequence A000045.
%F a(n+1) - a(n) = 1, 0 or -1.
%F Let F(n) = Fibonacci(n) and L(n) + Lucas (n).
%F For k >= 5, a(F(k) + j) = F(k-2) + j for 0 <= j <= F(k-2) (ascent to local peak value).
%F For k >= 3, a(L(k)) = 2*F(k-1) (local peak values).
%F For k >= 4, a(L(k) + j) = 2*F(k-1) - j, for 0 <= j <= F(k-3) (descent to trough).
%F For k >= 2, a(4*F(k) + j) = F(k+1) for 0 <= j <= F(k-3) (local trough values).
%e Sequence arranged to show local peak values P and troughs T:
%e 0,
%e 1,
%e 2,
%e 2,
%e 3,
%e P 4,
%e T 3,
%e 4,
%e 5,
%e P 6,
%e T 5, 5,
%e 6,
%e 7,
%e 8,
%e 9,
%e P 10,
%e 9,
%e T 8, 8,
%e 9,
%e 10,
%e 11,
%e 12,
%e 13,
%e 14,
%e 15,
%e P 16,
%e 15,
%e 14,
%e T 13, 13, 13,
%e 14,
%e 15,
%e 16,
%e 17,
%e 18,
%e 19,
%e 20,
%e 21,
%e 22,
%e 23,
%e 24,
%e 25,
%e P 26,
%e 25,
%e 24,
%e 23,
%e 22,
%e T 21, 21, 21, 21,
%e 22,
%e 23,
%e 24,
%e ...
%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)), n = 1..100);
%Y Cf. A000032, A000045, A356988, A356991 - A356999.
%K nonn,easy
%O 2,3
%A _Peter Bala_, Sep 11 2022
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