0, 0, 0, 1, 1, 2, 4, 8, 15, 29, 56, 108, 208, 401, 773, 1490, 2872, 5536, 10671, 20569, 39648, 76424, 147312, 283953, 547337, 1055026, 2033628, 3919944, 7555935, 14564533, 28074040, 54114452, 104308960, 201061985, 387559437, 747044834, 1439975216, 2775641472
Offset: 0
From _Petros Hadjicostas_, Mar 10 2018: (Start)
For n = 3, we get a(3+4) = a(7) = 8 palindromic compositions of 2*n+1 = 7 into an odd number of parts that are not a multiple of 4. They are the following: 7 = 1+5+1 = 3+1+3 = 2+3+2 = 1+2+1+2+1 = 2+1+1+1+2 = 1+1+3+1+1 = 1+1+1+1+1+1+1. If we put these compositions on a circle, they become bilaterally symmetric cyclic compositions of 2*n+1 = 7.
For n = 4, we get a(4+4) = a(8) = 15 palindromic compositions of 2*n + 1 = 9 into an odd number of parts that are not a multiple of 4. They are the following: 9 = 3+3+3 = 2+5+2 = 1+7+1 = 1+1+5+1+1 = 2+1+3+1+2 = 1+2+3+2+1 = 1+3+1+3+1 = 3+1+1+1+3 = 2+2+1+2+2 = 2+1+1+1+1+1+2 = 1+2+1+1+1+2+1 = 1+1+2+1+2+1+1 = 1+1+1+3+1+1+1 = 1+1+1+1+1+1+1+1+1.
As _David Callan_ points out in the comments above, for n >= 1, a(n+4) is also the number of 0-1 sequences of length n that avoid 1111. For example, for n = 5, a(5+4) = a(9) = 29 is the number of binary strings of length n that avoid 1111. Out of the 2^5 = 32 binary strings of length n = 5, the following do not avoid 1111: 11111, 01111, and 11110. (End)
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