A349802 Triangle read by rows: T(n,k) is the number of binary Lyndon words of length n that begin with exactly k 0's. 0 <= k <= n.
1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 2, 2, 1, 1, 0, 0, 2, 3, 2, 1, 1, 0, 0, 4, 6, 4, 2, 1, 1, 0, 0, 5, 10, 7, 4, 2, 1, 1, 0, 0, 8, 18, 14, 8, 4, 2, 1, 1, 0, 0, 11, 31, 26, 15, 8, 4, 2, 1, 1, 0, 0, 18, 56, 50, 30, 16, 8, 4, 2, 1, 1, 0
Offset: 0
Examples
For n = 6, the values correspond to the following Lyndon words: T(6,1) = 2 via 010111 and 011111; T(6,2) = 3 via 001011, 001101, and 001111; T(6,3) = 2 via 000101 and 000111; T(6,4) = 1 via 000011; and T(6,5) = 1 via 000001. Table begins: n\k | 0 1 2 3 4 5 6 7 8 9 ----+------------------------------------ 0 | 1 1 | 1, 1 2 | 0, 1, 0 3 | 0, 1, 1, 0 4 | 0, 1, 1, 1, 0 5 | 0, 2, 2, 1, 1, 0 6 | 0, 2, 3, 2, 1, 1, 0 7 | 0, 4, 6, 4, 2, 1, 1, 0 8 | 0, 5, 10, 7, 4, 2, 1, 1, 0 9 | 0, 8, 18, 14, 8, 4, 2, 1, 1, 0 ...
Links
- Andrew Howroyd, Table of n, a(n) for n = 0..1325 (rows n=0..50)
Programs
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PARI
B(k,n)=my(g=1/(1 - x*(1-x^k)/(1-x))); Vec(1 + sum(j=1, n, moebius(j)/j * log(subst(g + O(x*x^(n\j)), x, x^j)))) A(n,m)={my(M=Mat(vector(m, k, Col(B(k,n) - B(k-1,n))))); M[1,1]=M[2,2]=1; M} { my(M=A(10,10)); for(n=1, #M, print(M[n,1..n])) } \\ Andrew Howroyd, Dec 05 2021
Formula
T(n,n) = 0 for n >= 2.
T(n,n-1) = 1 for n >= 1.
T(n,n-m) = 2^(m-2) for n >= 2*m - 1 and m >= 2.
Comments