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 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. 2
 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, 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, 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS Not the characteristic function of A047564, as it does not contain 256, although here a(256) = 1. - Antti Karttunen, Oct 18 2018 LINKS Antti Karttunen, Table of n, a(n) for n = 0..65537 FORMULA 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 EXAMPLE a(2) = 0. So terms a(4) through a(7) are each equal to 1. a(3) = 1, so terms a(8) through a(15) are the same as terms a(0) through a(7). From Antti Karttunen, Oct 22 2018: (Start) For n = 256 = 2^8, a(8) = 0, thus a(256) = 1. For n = 2^64, a(64) = 0, thus a(2^64) = 1. (End) PROG (PARI) A000523(n) = if(n<1, 0, #binary(n)-1); A115971(n) = if(!n, n, if(!A115971(A000523(n)), 1, A115971(n-(2^A000523(n))))); \\ Antti Karttunen, Oct 18 2018 CROSSREFS Cf. A000523, A053645. Different from A320007. Sequence in context: A112690 A316343 A288864 * A320007 A072165 A072608 Adjacent sequences:  A115968 A115969 A115970 * A115972 A115973 A115974 KEYWORD easy,nonn AUTHOR Leroy Quet, Mar 14 2006 STATUS approved

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Last modified July 28 19:33 EDT 2021. Contains 346335 sequences. (Running on oeis4.)