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Total length of all longest runs of 0's in solus bitstrings of length n.
3

%I #22 Aug 21 2020 05:49:05

%S 0,1,4,9,18,34,62,110,192,331,565,958,1615,2710,4531,7552,12554,20823,

%T 34472,56972,94020,154959,255102,419532,689312,1131632,1856382,

%U 3043208,4985674,8163321,13359207,21851594,35726470,58386958,95383471,155766277,254288786

%N Total length of all longest runs of 0's in solus bitstrings of length n.

%C A bitstring is solus if all of its 1's are isolated.

%C The number of these bitstrings is A000045(n+2).

%H Alois P. Heinz, <a href="/A333394/b333394.txt">Table of n, a(n) for n = 0..1000</a>

%H Steven Finch, <a href="https://arxiv.org/abs/2003.09458">Cantor-solus and Cantor-multus distributions</a>, arXiv:2003.09458 [math.CO], 2020.

%F G.f.: Sum_{k>=1} (1+x)/(1-x-x^2)-(1+x-x^k-x^(k+1))/(1-x-x^2+x^(k+1)).

%e a(4) = 18 because the A000045(6) = 8 solus bitstrings of length 4 are 0000, 1000, 0100, 0010, 0001, 1010, 0101, 1001 and the longest 0-runs contribute 4+3+2+2+3+1+1+2 = 18.

%p b:= proc(n, w, m, s) option remember; `if`(n=0, m,

%p b(n-1, w+1, max(m, w+1), irem(s, 10)*10+0)+

%p `if`(s in [01, 21], 0, b(n-1, 0, m, irem(s, 10)*10+1)))

%p end:

%p a:= n-> b(n, 0, 0, 22):

%p seq(a(n), n=0..39); # _Alois P. Heinz_, Mar 18 2020

%t b[n_, w_, m_, s_] := b[n, w, m, s] = If[n == 0, m, b[n-1, w+1, Max[m, w+1], Mod[s, 10]*10+0]+If[MatchQ[s, 01|21], 0, b[n-1, 0, m, Mod[s, 10]*10+1]]];

%t a[n_] := b[n, 0, 0, 22];

%t a /@ Range[0, 39] (* _Jean-François Alcover_, Aug 21 2020, after _Alois P. Heinz_ *)

%Y Cf. A000045, A119706, A333395, A333396.

%K nonn

%O 0,3

%A _Steven Finch_, Mar 18 2020