A395191 - OEIS

%I #37 May 24 2026 16:53:05

%S 1,1,1,1,-1,1,1,1,1,1,1,-1,0,-1,1,1,1,0,-1,1,1,1,-1,1,0,-1,-1,1,1,1,2,

%T 2,1,0,1,1,1,-1,2,-3,2,-1,1,-1,1,1,1,1,3,4,2,0,1,1,1,1,-1,0,-2,4,-7,3,

%U 0,0,-1,1,1,1,0,0,1,7,10,3,1,-1,1,1,1,-1

%N Triangular array T(n, k) read by rows: Row n > 0 gives the coefficient of x^(n-k) in the expansion of f_n(x) = Sum_{k=0..n} T(n, k)*x^(n-k), where T(n, n) and T(0, n) = 1 for any n. The n-th falling diagonal has the ordinary generating function: 1/f_n(x).

%C The first falling diagonals are known to be periodic, the last periodic diagonal is starting at n=3, this causes periodicity in the first k columns up to k=3, this is the last periodic column.

%C k=6 is the last column that is a linear recurrence with constant coefficients.

%C The first difference between this triangle and A395188 occurs at T(7, 2), this is directly caused by row n=5 which is the first non-palindromic row in both triangles.

%F Sum_{k=0..n} T(n, k)*x^(n-k) = 1/Sum{m=0..oo} T(n+m, m)*x^m, for n > 0.

%F T(n, k) = Sum_{m=1..k} -T(n-m, k-m)*T(n-k, n-k-m), for n >= 2*k > 0.

%F T(n, 2) = 2*T(n-1, 2) - 2*T(n-2, 2) + T(n-3, 2), for n > 5.

%F T(n, 3) = -2*T(n-1, 3) - 3*T(n-2, 3) - 3*T(n-3, 3) - 2*T(n-4, 3) - T(n-5, 3), for n > 8.

%F T(n, 4) = 2*T(n-1, 4) - 3*T(n-2, 4) + 4*T(n-3, 4) - 4*T(n-4, 4) + 4*T(n-5, 4) - 3*T(n-6, 4) + 2*T(n-7, 4) - T(n-8, 4), for n > 12.

%F T(n, 5) = -3*T(n-1, 4) - 3*T(n-2, 5) - 2*T(n-3, 5) - 3*T(n-4, 5) - 5*T(n-5, 5) - 6*T(n-6, 5) - 6*T(n-7, 5) - 6*T(n-8, 5) - 6*T(n-9, 5) - 6*T(n-10, 5) - 6*T(n-11, 5) - 5*T(n-12, 5) - 3*T(n-13, 5) - 3*T(n-14, 5) - 4*T(n-15, 5) - 3*T(n-16, 5) - T(n-17, 5), for n > 22.

%F T(n, 6) = 3*T(n-1, 6) - 2*T(n-2, 6) - T(n-3, 6) + 5*T(n-5, 6) - 8*T(n-6, 6) + 4*T(n-7, 6) + T(n-9, 6) - 4*T(n-10, 6) + 5*T(n-11, 6) - 2*T(n-12, 6) - 2*T(n-13, 6) + 2*T(n-14, 6) + T(n-15, 6) - 5*T(n-17, 6) + 8*T(n-18, 6) - 4*T(n-19, 6) - T(n-21, 6) + 4*T(n-22, 6) - 5*T(n-23, 6) + 3*T(n-24, 6) - T(n-25, 6), for n > 31.

%e Triangle T(n, k) starts:

%e   [ 0] 1;

%e   [ 1] 1,  1;

%e   [ 2] 1, -1,  1;

%e   [ 3] 1,  1,  1,  1;

%e   [ 4] 1, -1,  0, -1,  1;

%e   [ 5] 1,  1,  0, -1,  1,  1;

%e   [ 6] 1, -1,  1,  0, -1, -1,  1;

%e   [ 7] 1,  1,  2,  2,  1,  0,  1,  1;

%e   [ 8] 1, -1,  2, -3,  2, -1,  1, -1,  1;

%e   [ 9] 1,  1,  1,  3,  4,  2,  0,  1,  1,  1;

%e   [10] 1, -1,  0, -2,  4, -7,  3,  0,  0, -1,  1;

%e   ...

%e The falling diagonal beginning in row n = 2 is: 1, 1, 0, -1, -1, 0, 1, ... . It has the generating function: 1/(x^2-x+1). The coefficients of x^2-x+1 are 1, -1, 1, this is row n = 2.

%o (PARI)

%o squareRow(n, max_k) = if(n==0, vector(max_k, k, 1), my(f(x)=x^n+sum(k=1, n, squareRow(n-k, k+1)[k+1]*x^(n-k))); Vec(1/f(x)+O(x^max_k)))

%o T(n, k) = squareRow(n-k, k+1)[k+1]

%Y Cf. A395188 (same idea but reversed order of terms relation in rows).

%K sign,easy,tabl

%O 0,31

%A _Thomas Scheuerle_, Apr 15 2026