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

1,3

Links

Table of n, a(n) for n=1..105.

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

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

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

Cf. A000002, A071928, A229785, A284796.

Sequence in context: A308884 A101598 A342461 * A101615 A246359 A328114

Adjacent sequences: A157126 A157127 A157128 * A157130 A157131 A157132

Keyword

nonn

Author

Benoit Cloitre, Feb 23 2009

Status

approved