A243815 Number of length n words on alphabet {0,1} such that the length of every maximal block of 0's (runs) is the same.
1, 2, 4, 8, 14, 24, 39, 62, 97, 151, 233, 360, 557, 864, 1344, 2099, 3290, 5176, 8169, 12931, 20524, 32654, 52060, 83149, 133012, 213069, 341718, 548614, 881572, 1417722, 2281517, 3673830, 5918958, 9540577, 15384490, 24817031, 40045768, 64637963, 104358789
Offset: 0
Keywords
Examples
0110 is a "good" word because the length of both its runs of 0's is 1. Words of the form 11...1 are good words because the condition is vacuously satisfied. a(5) = 24 because there are 32 length 5 binary words but we do not count: 00010, 00101, 00110, 01000, 01001, 01100, 10010, 10100.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
Programs
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Maple
a:= n-> 1 + add(add((d-> binomial(d+j, d))(n-(i*j-1)) , j=1..iquo(n+1, i)), i=2..n+1): seq(a(n), n=0..50); # Alois P. Heinz, Jun 11 2014
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Mathematica
nn=30;Prepend[Map[Total,Transpose[Table[Drop[CoefficientList[Series[ (1+x^k)/(1-x-x^(k+1))-1/(1-x),{x,0,nn}],x],1],{k,1,nn}]]],0]+1 Table[Length[Select[Subsets[Range[n]],SameQ@@Length/@Split[#,#2==#1+1&]&]],{n,0,10}] (* Gus Wiseman, Jun 23 2025 *)
Comments