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A094186
Taking a(1)=0 and a(2)=1, sequence (a(n))n>1 is defined as follows : letting w(k)=a(1)a(2)...a(k) and w(infinity)= limit k ->infinity a(1)a(2)w(1)w(2)...w(k) we have w(infinity)=a(1)a(2)a(3)a(4)...
7
0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0
OFFSET
1,1
COMMENTS
An infinite binary word.
A shorter definition: the limit of the string "0, 1" under the operation 'append first k terms, increment k' with k=1 initially.
Sums of the first 10^n terms are: 0, 4, 36, 358, 3576, 34908, 356258, 3621799, 35807401, 352047694, 3495167093. [Alex Ratushnyak, Aug 15 2012]
LINKS
FORMULA
It seems that limit n ->infinity 1/n*sum(k=1, n, a(k)) = 0.34...
EXAMPLE
w(1)=0, w(2)=01, therefore a(1)a(2)w(1)w(2)=01001=a(1)a(2)a(3)a(4)a(5) and sequence begins : 0,1,0,0,1,...
MAPLE
S:= "01":
for k from 1 to 40 do
A:= cat(A, A[1..k])
od:
seq(parse(A[i]), i=1..length(A)); # Robert Israel, Mar 28 2019
PROG
(Python)
TOP = 1000
a = [0]*TOP
a[1] = 1
n = 2
k = 1
while n+k < TOP:
a[n:] = a[:k]
n += k
k += 1
for k in range(n):
print(a[k], end=", ")
# Alex Ratushnyak, Aug 15 2012
CROSSREFS
KEYWORD
nonn
AUTHOR
Benoit Cloitre, May 07 2004
STATUS
approved