%I #9 Aug 24 2019 02:26:52
%S 1,1,1,1,2,1,1,1,3,1,1,2,3,4,1,1,1,1,6,5,1,1,2,3,4,10,6,1,1,1,3,5,10,
%T 15,7,1,1,2,1,4,10,20,21,8,1,1,1,3,6,11,21,35,28,9,1,1,2,3,4,10,24,42,
%U 56,36,10,1,1,1,1,5,10,25,49,78,84,45,11,1
%N Triangle read by rows: T(n,k) is the number of compositions of n with k parts and circular differences all equal to 1, 0, or -1.
%F T(n, 1) = T(n, n) = 1.
%F T(n, 2) = (3 - (-1)^n)/2.
%F T(n, n - 1) = binomial(n-1, 1) = n - 1.
%F T(n, n - 2) = binomial(n-2, 2).
%e Triangle begins:
%e 1;
%e 1, 1;
%e 1, 2, 1;
%e 1, 1, 3, 1;
%e 1, 2, 3, 4, 1;
%e 1, 1, 1, 6, 5, 1;
%e 1, 2, 3, 4, 10, 6, 1;
%e 1, 1, 3, 5, 10, 15, 7, 1;
%e 1, 2, 1, 4, 10, 20, 21, 8, 1;
%e 1, 1, 3, 6, 11, 21, 35, 28, 9, 1;
%e 1, 2, 3, 4, 10, 24, 42, 56, 36, 10, 1;
%e 1, 1, 1, 5, 10, 25, 49, 78, 84, 45, 11, 1;
%e 1, 2, 3, 4, 10, 24, 56, 96, 135, 120, 55, 12, 1;
%e 1, 1, 3, 6, 10, 21, 57, 116, 180, 220, 165, 66, 13, 1;
%e ...
%e For n = 6 there are a total of 15 compositions:
%e k = 1: (6)
%e k = 2: (33)
%e k = 3: (222)
%e k = 4: (1122), (1212), (1221), (2112), (2121), (2211)
%e k = 5: (11112), (11121), (11211), (12111), (21111)
%e k = 6: (111111)
%o (PARI)
%o step(R,n)={matrix(n, n, i, j, if(i>j, if(j>1, R[i-j, j-1]) + R[i-j, j] + if(j+1<=n, R[i-j, j+1])) )}
%o T(n)={my(v=vector(n)); for(k=1, n, my(R=matrix(n, n, i, j, i==j&&abs(i-k)<=1), m=0); while(R, m++; v[m]+=R[n, k]; R=step(R, n))); v}
%o for(n=1, 12, print(T(n)));
%Y Row sums are A325591.
%Y Cf. A309937, A309938, A309939.
%K nonn,tabl
%O 1,5
%A _Andrew Howroyd_, Aug 23 2019