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A157129
An analog of the Kolakoski sequence A000002, only now a(n) = (length of n-th run divided by 2) using 1 and 2 and starting with 1,1.
2
1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 2
OFFSET
1,3
FORMULA
As for the Kolakoski sequence we suspect Sum_{k=1..n} a(k) = (3/2)*n + o(n).
a(n) = A071928(n)/2. - Jon Maiga, Jun 04 2021
a(n) = gcd(A284796(ceiling(n/2)), 2) (conjectured). - Jon Maiga, Jun 11 2021
EXAMPLE
The third run is 1,1,1,1, which is of length 4, thus a(3) = 4/2 = 2.
PROG
(PARI) w=[1, 1]; for(n=2, 1000, for(i=1, 2*w[n], w=concat(w, 1+(n+1)%2))); w \\ Corrected by Kevin Ryde and Jon Maiga, Jun 11 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Feb 23 2009
STATUS
approved