%I #39 Jun 05 2024 21:53:20
%S 0,0,0,1,0,1,1,0,0,1,4,2,4,2,4,3,0,0,0,0,3,40,36,40,40,40,36,40,29,0,
%T 0,0,0,0,0,29,232,152,240,200,208,200,240,152,232,235,142,140,257,168,
%U 168,257,140,142,235,11712,13216,12208,12384,11408,11136,11408,12384,12208,13216,11712
%N Triangle read by rows: T(n,k) = number of permutations of [n] starting from k that have zero (n-1)-th differences. (n>=1, 1<=k<=n).
%H Seiichi Manyama, <a href="/A373422/b373422.txt">Rows n = 1..14, flattened</a>
%F T(n,k) = T(n,n+1-k) for 1<=k<=n.
%F If p is prime, T(p+1,k) = 0 for 2 <= k <= p.
%e T(3,1) = 1 because [1,2,3] have zero 2nd differences.
%e 1 2 3
%e 1 1
%e 0
%e Triangle starts:
%e 0;
%e 0, 0;
%e 1, 0, 1;
%e 1, 0, 0, 1;
%e 4, 2, 4, 2, 4;
%e 3, 0, 0, 0, 0, 3;
%e 40, 36, 40, 40, 40, 36, 40;
%e 29, 0, 0, 0, 0, 0, 0, 29;
%e 232, 152, 240, 200, 208, 200, 240, 152, 232;
%e 235, 142, 140, 257, 168, 168, 257, 140, 142, 235;
%o (PARI) tabl(n) = my(nn=vector(n)); forperm([1..n], p, if(sum(k=1, n, (-1)^k*binomial(n-1, k-1)*p[k])==0, nn[p[1]]++)); nn;
%Y Row sums give 2 * A131502(n-1).
%K nonn,tabl
%O 1,11
%A _Seiichi Manyama_, Jun 04 2024