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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.

%I A243815 #24 Jun 26 2025 18:56:08
%S A243815 1,2,4,8,14,24,39,62,97,151,233,360,557,864,1344,2099,3290,5176,8169,
%T A243815 12931,20524,32654,52060,83149,133012,213069,341718,548614,881572,
%U A243815 1417722,2281517,3673830,5918958,9540577,15384490,24817031,40045768,64637963,104358789
%N 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.
%C A243815 Number of terms of A164710 with exactly n+1 binary digits. - _Robert Israel_, Nov 09 2015
%C A243815 From _Gus Wiseman_, Jun 23 2025: (Start)
%C A243815 This is the number of subsets of {1..n} with all equal lengths of runs of consecutive elements increasing by 1. For example, the runs of S = {1,2,5,6,8,9} are ((1,2),(5,6),(8,9)), with lengths (2,2,2), so S is counted under a(9). The a(0) = 1 through a(4) = 14 subsets are:
%C A243815   {}  {}   {}     {}       {}
%C A243815       {1}  {1}    {1}      {1}
%C A243815            {2}    {2}      {2}
%C A243815            {1,2}  {3}      {3}
%C A243815                   {1,2}    {4}
%C A243815                   {1,3}    {1,2}
%C A243815                   {2,3}    {1,3}
%C A243815                   {1,2,3}  {1,4}
%C A243815                            {2,3}
%C A243815                            {2,4}
%C A243815                            {3,4}
%C A243815                            {1,2,3}
%C A243815                            {2,3,4}
%C A243815                            {1,2,3,4}
%C A243815 (End)
%H A243815 Alois P. Heinz, <a href="/A243815/b243815.txt">Table of n, a(n) for n = 0..1000</a>
%e A243815 0110 is a "good" word because the length of both its runs of 0's is 1.
%e A243815 Words of the form 11...1 are good words because the condition is vacuously satisfied.
%e A243815 a(5) = 24 because there are 32 length 5 binary words but we do not count: 00010, 00101, 00110, 01000, 01001, 01100, 10010, 10100.
%p A243815 a:= n-> 1 + add(add((d-> binomial(d+j, d))(n-(i*j-1))
%p A243815           , j=1..iquo(n+1, i)), i=2..n+1):
%p A243815 seq(a(n), n=0..50);  # _Alois P. Heinz_, Jun 11 2014
%t A243815 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
%t A243815 Table[Length[Select[Subsets[Range[n]],SameQ@@Length/@Split[#,#2==#1+1&]&]],{n,0,10}] (* _Gus Wiseman_, Jun 23 2025 *)
%Y A243815 Cf. A164710.
%Y A243815 These subsets are ranked by A164707, complement A164708.
%Y A243815 For distinct instead of equal lengths we have A384175, complement A384176.
%Y A243815 For anti-runs instead of runs we have A384889, for partitions A384888.
%Y A243815 For permutations instead of subsets we have A384892, distinct instead of equal A384891.
%Y A243815 For partitions instead of subsets we have A384904, strict A384886.
%Y A243815 The complement is counted by A385214.
%Y A243815 A034839 counts subsets by number of maximal runs, for strict partitions A116674.
%Y A243815 A049988 counts partitions with equal run-lengths, distinct A325325.
%Y A243815 A329738 counts compositions with equal run-lengths, distinct A329739.
%Y A243815 A384887 counts partitions with equal lengths of gapless runs, distinct A384884.
%Y A243815 Cf. A010027, A044813, A069010, A268193, A383013, A384177, A384893.
%K A243815 nonn
%O A243815 0,2
%A A243815 _Geoffrey Critzer_, Jun 11 2014