OFFSET
1,3
COMMENTS
We conjecture that the density of 1's in the sequence approaches 2/3 as n -> infinity. This conjecture is proved in the paper of Shallit.
From Daniel Hoyt, Sep 05 2025: (Start)
This sequence lists the run lengths of the first absolute differences of the dragon curve, the regular paper-folding sequence (A014577).
The blocks of (1, 1) and (2) of this sequence interleave central binomial coefficients in the continued fraction of Sum_{k>=0} 1/(2^k)! (see A387398). (End)
From Daniel Hoyt, Nov 21 2025: (Start)
A paperfolding construction: Use the binary alphabet (11, 2), where the symbol "11" stands for two consecutive 1's and "2" stands for a single 2. For a word W, let rev(W) be W reversed. Let revflip(W) be rev(W) with its first symbol flipped (11 <-> 2). Let "+" be used for concatenation.
Seed: A0 = 11.
Recursive step: A_{k+1} = A_k + revflip(A_k).
Examples:
A0 = 11
A1 = 112 (rev(11)=11; flip first 11->2; append 2)
A2 = 1121111 (rev(112)=211; flip first 2->11; append 1111)
A3 = 1121111211211 (rev(1121111)=1111211; flip first 11->2; append 211211)
A4 = 1121111211211221121111211 (and so on)
From A3 onward the first symbol of rev(A_k) is always "11", so the flip is always 11->2. The limit of A_k is exactly this sequence. (End)
LINKS
Benoit Cloitre and Jeffrey Shallit, A Self-Generating Sequence, Theoret. Comp. Sci. (2025) 115721.
Daniel Hoyt, Representing the sequence visually.
Daniel Hoyt, Vector drawing web app of the visual representation.
Daniel Hoyt, The Self-Describing Paperfolding Sequence in Continued Fractions, J. Int. Seq. 29 (2026), Art. 26.3.6.
Jeffrey Shallit, Cloitre's Self-Generating Sequence, arXiv:2501.00784 [math.CO], 2025.
EXAMPLE
Write the sums of elements in each run, you obtain: 2,2,4,2,2,2,2,4,2,2,4,2,2,4,4,... dividing by 2 you get: 1,1,2,1,1,1,1,2,1,1,2,1,1,2,2,... the sequence itself.
MAPLE
mx:= 1000: l:= [1$2]: a:= n-> l[n]:
for h from 2 while nops(l)<mx do
t:= 2-irem(h, 2); l:= [l[], t$(l[h]*2/t)]
od:
seq (a(n), n=1..120); # Alois P. Heinz, May 31 2012
PROG
(Python)
def a157196(n):
blk = ['11'] # A_0
while sum(map(len, blk)) < n: # grow until we have >= n digits
b = list(reversed(blk))
b[0] = '11' if b[0] == '2' else '2' # flip first block of the mirror
blk += b
print(', '.join(''.join(blk)[:n]))
a157196(100) # Number of digits
CROSSREFS
KEYWORD
nonn,nice
AUTHOR
Benoit Cloitre, Feb 24 2009
EXTENSIONS
More terms from Alvin Hoover Belt, May 31 2012
STATUS
approved