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%I A115971 #16 Oct 22 2018 12:44:50
%S A115971 0,1,0,1,1,1,1,1,0,1,0,1,1,1,1,1,0,1,0,1,1,1,1,1,0,1,0,1,1,1,1,1,0,1,
%T A115971 0,1,1,1,1,1,0,1,0,1,1,1,1,1,0,1,0,1,1,1,1,1,0,1,0,1,1,1,1,1,0,1,0,1,
%U A115971 1,1,1,1,0,1,0,1,1,1,1,1,0,1,0,1,1,1,1,1,0,1,0,1,1,1,1,1,0,1,0,1,1,1,1,1,0
%N A115971 a(0) = 0. If a(n) = 0, then a(2^n) through a(2^(n+1)-1) are each equal to 1. If a(n) = 1, then a(m + 2^n) = a(m) for each m, 0 <= m <= 2^n -1.
%C A115971 Not the characteristic function of A047564, as it does not contain 256, although here a(256) = 1. - _Antti Karttunen_, Oct 18 2018
%H A115971 Antti Karttunen, <a href="/A115971/b115971.txt">Table of n, a(n) for n = 0..65537</a>
%H A115971 <a href="/index/Ch#char_fns">Index entries for characteristic functions</a>
%F A115971 a(0) = 0; and for n > 0, if a(A000523(n)) is 0, then a(n) = 1, otherwise a(n) = a(n-(2^A000523(n))) = a(A053645(n)). - _Antti Karttunen_, Oct 22 2018
%e A115971 a(2) = 0. So terms a(4) through a(7) are each equal to 1.
%e A115971 a(3) = 1, so terms a(8) through a(15) are the same as terms a(0) through a(7).
%e A115971 From _Antti Karttunen_, Oct 22 2018: (Start)
%e A115971 For n = 256 = 2^8, a(8) = 0, thus a(256) = 1.
%e A115971 For n = 2^64, a(64) = 0, thus a(2^64) = 1. (End)
%o A115971 (PARI)
%o A115971 A000523(n) = if(n<1, 0, #binary(n)-1);
%o A115971 A115971(n) = if(!n,n,if(!A115971(A000523(n)),1,A115971(n-(2^A000523(n))))); \\ _Antti Karttunen_, Oct 18 2018
%Y A115971 Cf. A000523, A053645.
%Y A115971 Different from A320007.
%K A115971 easy,nonn
%O A115971 0,1
%A A115971 _Leroy Quet_, Mar 14 2006

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