1, 2, 5, 12, 28, 64, 144, 320, 704, 1536, 3328, 7168, 15360, 32768, 69632, 147456, 311296, 655360, 1376256, 2883584, 6029312, 12582912, 26214400, 54525952, 113246208, 234881024, 486539264, 1006632960, 2080374784, 4294967296, 8858370048, 18253611008, 37580963840
Offset: 0
E.g. a(2)=5 because in the compositions of 3, namely 3,2+1,1+2,1+1+1, we have five 1's altogether.
There are a(3)=12 compositions of 3 into 2 sorts of parts where all parts of the first sort precede all parts of the second sort. Here p:s stands for "part p of sort s":
01: [ 1:0 1:0 1:0 ]
02: [ 1:0 1:0 1:1 ]
03: [ 1:0 1:1 1:1 ]
04: [ 1:0 2:0 ]
05: [ 1:0 2:1 ]
06: [ 1:1 1:1 1:1 ]
07: [ 1:1 2:1 ]
08: [ 2:0 1:0 ]
09: [ 2:0 1:1 ]
10: [ 2:1 1:1 ]
11: [ 3:0 ]
12: [ 3:1 ]
- _Joerg Arndt_, Apr 28 2013
For the compositions of 6, the total number of runs of parts of size 2 is a(6-2) - a(6-2*2) = 28 - 5 = 23, enumerated as follows (with the runs of 2 enclosed in []): 4,[2]; [2],4; [2],3,1; [2],1,3; 3,[2],1; 1,[2],3; 3,1,[2]; 1,3,[2]; [2,2,2]; [2,2],1,1; 1,[2,2],1; 1,1,[2,2]; [2],1,[2],1; 1,[2],1,[2]; [2],1,1,[2]; [2],1,1,1,1; 1,[2],1,1,1; 1,1,[2],1,1; 1,1,1,[2],1; and 1,1,1,1[2]. - _Gregory L. Simay_, Feb 17 2018
There are a(3)=12 triwords of length 3: (0,0,0), (0,0,2), (0,2,0), (0,2,2), (1,0,0), (1,0,2), (1,1,0), (1,1,1), (1,1,2), (1,2,0), (1,2,1), (1,2,2). - _Henri Mühle_, Mar 08 2021
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