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A244229 a(n) = Number of integers 0 < k <= n, which have an even representation in Greedy Catalan Base (A014418).

%I A244229 #12 Jun 25 2014 13:12:17
%S A244229 0,0,1,1,2,3,3,4,4,5,6,6,7,7,8,8,9,9,10,11,11,12,12,13,14,14,15,15,16,
%T A244229 16,17,17,18,19,19,20,20,21,22,22,23,23,24,24,25,25,26,27,27,28,28,29,
%U A244229 30,30,31,31,32,32,33,33,34,35,35,36,36,37,38,38,39,39,40,40,41,41,42
%N A244229 a(n) = Number of integers 0 < k <= n, which have an even representation in Greedy Catalan Base (A014418).
%C A244229 This works as an inverse function for the injection A244222. We have a(A244222(n)) = n for all n.
%C A244229 Equally, for n >= 0, a(n) = the largest k such that A244222(k) <= n.
%H A244229 Antti Karttunen, <a href="/A244229/b244229.txt">Table of n, a(n) for n = 0..4862</a>
%F A244229 a(n) = A244224(n)-1.
%F A244229 a(n) = n - A244225(n).
%e A244229 The first positive numbers in Greedy Catalan Base representation are:
%e A244229 A014418(1) = 1
%e A244229 A014418(2) = 10
%e A244229 A014418(3) = 11
%e A244229 A014418(4) = 20
%e A244229 A014418(5) = 100
%e A244229 A014418(6) = 101
%e A244229 A014418(7) = 110
%e A244229 Of these, the first "even" representation (ending with zero) occurs at n=2, thus a(0) = a(1) = 0, and a(2) = 1. The next even representations occur at n=4, 5 and 7, thus a(3) = 1, a(4) = 2, a(5) = 3, a(6) = 3 and a(7) = 4.
%o A244229 (Scheme, two alternative definitions)
%o A244229 (define (A244229 n) (- (A244224 n) 1))
%o A244229 (define (A244229 n) (- n (A244225 n)))
%Y A244229 One less than A244224 (partial sums of A244220).
%Y A244229 Cf. A014418, A244220, A244222, A244225.
%K A244229 nonn
%O A244229 0,5
%A A244229 _Antti Karttunen_, Jun 25 2014