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a(n) = a(a(a(a(a(n-2))))) + a(n - a(n-2)).
1

%I #17 Aug 03 2023 01:42:42

%S 1,1,2,3,3,3,4,4,4,4,5,5,5,5,5,6,6,6,6,6,6,7,7,7,7,7,7,7,8,8,8,8,8,8,

%T 8,8,9,9,9,9,9,9,9,9,9,10,10,10,10,10,10,10,10,10,10,11,11,11,11,11,

%U 11,11,11,11,11,11,12,12,12,12,12,12,12,12,12,12,12,12,13,13,13,13,13,13,13

%N a(n) = a(a(a(a(a(n-2))))) + a(n - a(n-2)).

%C A fifth-order recursion based on A005229.

%C Differs from A106742 at n = 233 where this is 23 whereas A106742(233) = 22. - _Joshua Zucker_, May 17 2007

%t a[n_Integer?Positive] := a[n] = a[a[a[a[a[n-2]]]]]+ a[n - a[n-2]]

%t a[1] = a[2] = 1

%t Table[a[n], {n, 1, 100}]

%Y Cf. A004001, A005229, A106742.

%K nonn

%O 1,3

%A _Roger L. Bagula_, Oct 07 2003

%E Edited by _N. J. A. Sloane_, Jun 15 2007