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Sum of ceiling(n/per(w)) over all binary words of length n.
4

%I #19 Feb 07 2020 11:23:58

%S 2,6,14,30,60,118,236,460,914,1810,3608,7158,14310,28504,56978,113778,

%T 227484,454534,909050,1817232,3634344,7267198,14534120,29064982,

%U 58129922,116253394,232506236,465000468,929999880,1859974762,3719949488,7439848936,14879695742

%N Sum of ceiling(n/per(w)) over all binary words of length n.

%C The period per(w), for w = w[1..n] a word, is the least p >= 1 such that w[i] = w[i+p] for 1 <= i <= n-p.

%C Asymptotically we have a(n) ~ 1.732213...*2^n.

%H D. Gabric and J. Shallit, <a href="https://arxiv.org/abs/2002.01968">Avoidance of split overlaps</a>, arxiv preprint arXiv:2002.01968 [cs.DM], February 5 2020.

%H Rémy Sigrist, <a href="/A331699/a331699.txt">C program for A331699</a>, Jan 26 2020.

%e For n = 3 there are two words of period 1 (000 and 111), two words of period 2 (010 and 101), and all other words are of period 3. So a(n) = 2*ceiling(3/1) + 2*ceiling(3/2) + 4*ceiling(3/3) = 14.

%o (C) See Links section.

%K nonn

%O 1,1

%A _Jeffrey Shallit_, Jan 25 2020

%E a(25)-a(33) from _Rémy Sigrist_, Jan 26 2020