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.
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
Keywords
Examples
The third run is 1,1,1,1, which is of length 4, thus a(3) = 4/2 = 2.
Programs
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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
Formula
As for the Kolakoski sequence we suspect Sum_{k=1..n} a(k) = (3/2)*n + o(n).
Generated by infinitely iterating the morphism a->abc, b->dab, c->efg, d->hcd, e->cda, f->bef, g->ghc, h->dab starting with a, obtaining the infinite word abcdabefg..., and then replacing a,b,e,f by 1 and c,d,g,h by 2. Using Walnut, one can then prove the above claim about Sum_{k=1..n} a(k) in the stronger form Sum_{k=1..n} a(k) = (3/2)*n + O(1). Jeffrey Shallit, Dec 31 2024