

A287775


Positions of 0 in A287772; complement of A050140 (conjectured and proved).


4



2, 3, 6, 7, 9, 10, 13, 14, 17, 18, 20, 21, 24, 25, 27, 28, 31, 32, 35, 36, 38, 39, 42, 43, 46, 47, 49, 50, 53, 54, 56, 57, 60, 61, 64, 65, 67, 68, 71, 72, 74, 75, 78, 79, 82, 83, 85, 86, 89, 90, 93, 94, 96, 97, 100, 101, 103, 104, 107, 108, 111, 112, 114
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OFFSET

1,1


COMMENTS

1 < n*r  a(n) < 1 for n >= 1, where r = (5 + sqrt(5))/4.
From Michel Dekking, Dec 28 2017: (Start)
Let (d(n)) be the sequence of first differences: d(n)=a(n+1)a(n).
CLAIM: d(n) = A108103(n+1) for n=1,2,….
Proof: As a word A287772 = 100110010011001100100110010011... obtained by substituting 0>1, 1>00 in the Fibonacci word F=0100101001001010010100...
This implies that A287772 is a concatenation of 00’s separated by 1’s and 11’s. Moreover, a 0110 occurs iff 1001 occurs in F, and a 010 occurs iff 101 occurs in F. Note also that occurrence of a 00 in A287772 yields a d(n)=1 (and so every other letter in d is a 1), occurrence of a 010 yields a d(n)=2, and occurrence of a 0110 yields a d(n)=3. Since the 1001’s and 101’s occur in 1F according to F itself with 1 prepended (see A001468 and A282162), we must have d(n)=A108103(n+1). (End)


LINKS

Clark Kimberling, Table of n, a(n) for n = 1..10000


MATHEMATICA

s = Nest[Flatten[# /. {0 > {0, 1}, 1 > {0}}] &, {0}, 10] (* A003849 *)
w = StringJoin[Map[ToString, s]]
w1 = StringReplace[w, {"0" > "1", "1" > "00"}]
st = ToCharacterCode[w1]  48 (* A287772 *)
Flatten[Position[st, 0]] (* A287775 *)
Flatten[Position[st, 1]] (* A050140 conjectured *)


CROSSREFS

Cf. A050140, A287772, A108103.
Sequence in context: A191215 A190847 A201734 * A283208 A259726 A259587
Adjacent sequences: A287772 A287773 A287774 * A287776 A287777 A287778


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, Jun 03 2017


STATUS

approved



