A349802 - OEIS

%I #31 Dec 06 2021 12:36:39

%S 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,

%T 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,

%U 0,0,18,56,50,30,16,8,4,2,1,1,0

%N 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.

%C Rows sum to A001037.

%C Conjecture: The Euler transform of column k=1 gives the Fibonacci numbers, the Euler transform of column k=2 gives the tribonacci numbers (A000073), and more generally, the Euler transform of column k >= 1 gives the (k+1)-bonacci numbers (A048887).

%H Andrew Howroyd, <a href="/A349802/b349802.txt">Table of n, a(n) for n = 0..1325</a> (rows n=0..50)

%F T(n,n) = 0 for n >= 2.

%F T(n,n-1) = 1 for n >= 1.

%F T(n,n-m) = 2^(m-2) for n >= 2*m - 1 and m >= 2.

%e For n = 6, the values correspond to the following Lyndon words:

%e T(6,1) = 2 via 010111 and 011111;

%e T(6,2) = 3 via 001011, 001101, and 001111;

%e T(6,3) = 2 via 000101 and 000111;

%e T(6,4) = 1 via 000011; and

%e T(6,5) = 1 via 000001.

%e Table begins:

%e n\k | 0  1   2   3  4  5  6  7  8  9

%e ----+------------------------------------

%e   0 | 1

%e   1 | 1, 1

%e   2 | 0, 1,  0

%e   3 | 0, 1,  1,  0

%e   4 | 0, 1,  1,  1, 0

%e   5 | 0, 2,  2,  1, 1, 0

%e   6 | 0, 2,  3,  2, 1, 1, 0

%e   7 | 0, 4,  6,  4, 2, 1, 1, 0

%e   8 | 0, 5, 10,  7, 4, 2, 1, 1, 0

%e   9 | 0, 8, 18, 14, 8, 4, 2, 1, 1, 0

%e   ...

%o (PARI)

%o 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))))

%o 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}

%o { my(M=A(10,10)); for(n=1, #M, print(M[n,1..n])) } \\ _Andrew Howroyd_, Dec 05 2021

%Y Cf. A000045, A000073, A001037, A006206 (k=1), A048887.

%K nonn,tabl

%O 0,17

%A _Peter Kagey_, Nov 30 2021