OFFSET
1,3
COMMENTS
Equivalently, a(n) = 2*(common value of n-th run)/(length of n-th run) using 1 and 2 starting with 1,1.
A self-describing paperfolding sequence obtained from the regular paperfolding sequence. It is the dual of A157196, having the opposite limiting ratio of 1's to 2's, and can be formed from its run lengths r(n) = A379704(n) by omitting r(2), and expanding r(1), r(3), r(4), r(5), ... into alternating runs of 1 and 2.
A paperfolding construction: Use the binary alphabet (1, 22), where the symbol "1" stands for a single 1 and "22" stands for two consecutive 2's. For a word W, let rev(W) be W reversed. Let revendflip(W) be rev(W) with both its first and last symbols flipped (1 <-> 22). Let + be used for concatenation.
Seed: A_0 = 11.
Recursive step: A_{k+1} = A_k + revendflip(A_k).
Examples:
A_0 = 11
A_1 = 112222 (rev(11)=11; flip first 1->22 and last 1->22; append 2222)
A_2 = 112222122122 (rev(112222)=222211; flip first 22->1 and last 1->22; append 122122)
A_3 = 112222122122112212222122 (and so on)
The limit of A_k is this sequence.
To demonstrate the relationship between A157196, A395159, and the regular paperfolding sequence A014577: write A157196 in pieces 11 and 2, and write A395159 in pieces 1 and 22. Omit the first piece from each sequence, namely the initial 11 from A157196 and the initial 1 from A395159. Then pair the remaining pieces:
A157196: 2 | 11 | 11 | 2 | 11 | 2 | 11 | 2 | 2 | 11 | ...
A395159: 1 | 22 | 22 | 1 | 22 | 1 | 22 | 1 | 1 | 22 | ...
Read single digits as "repeat the previous bit", and double digits as "change the previous bit". Starting with bit 1 gives 1,1,0,1,1,0,0,1,1,1,0,0,1,0,0,1,... which is A014577.
LINKS
EXAMPLE
The maximal runs are
11 | 2222 | 1 | 22 | 1 | 22 | 11 | 22 | 1 | 2222 | ...
Let S be the sum of the elements of the n-th run, and L be the length of the n-th run. Then, for these runs, the values 2*S/L^2 are 1, 1, 2, 2, 2, 2, 1, 2, 2, 1, ... which are the terms of the sequence.
PROG
(Python)
#a(n), b(n), and c(n) are three equivalent ways to generate this sequence.
def a(n): #Demonstrating a(n) = 2*(common value of n-th run)/(length of n-th run)
s=[1, 1]; r=2
while len(s)<n:
v=2-r%2; s+=[v]*int(2*v/s[r-1]); r+=1
return s[:n]
def b(n): #Demonstrating b(n) = 2*(sum of elements of n-th run)/(length of n-th run)^2
s=[1, 1]; r=2
while len(s)<n:
v=2-r%2; L=1
while True:
S=sum([v]*L); N=2*S; D=L**2; q=int(N/D)
if q==s[r-1] and q*D==N: break
L+=1
s+=[v]*L; r+=1
return s[:n]
def c(n): #Demonstrating the folding construction.
w=[1, 1]
k=0 if n<=2 else (((n+5)//6)-1).bit_length()+1
for _ in range(k):
r=w[::-1]; r[0]=3-r[0]; r[-1]=3-r[-1]; w+=r
return [y for x in w for y in [x]*x][:n]
CROSSREFS
KEYWORD
nonn,new
AUTHOR
Daniel Hoyt, Jun 18 2026
STATUS
approved