A395188 - OEIS

%I #45 May 24 2026 16:58:31

%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,-1,1,1,1,1,

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

%U 0,0,-1,1,1,1,1,1,2,3,-3,3,1,-1,1,1,1,-1

%N Triangular array T(n, k) read by rows: Row n > 0 gives the coefficient of x^k in the expansion of f_n(x) = Sum_{k=0..n} T(n, k)*x^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 columns up to k = 5 become periodic if we omit the first term. However column k = 6 is not periodic anymore and is also the last column known as a linear recurrence with constant coefficients.

%C The first difference between this triangle and A395191 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^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, m), for n > k.

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

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

%F T(n, 4) = -T(n-2, 4) + T(n-3, 4) - T(n-4, 4) + T(n-5, 4) - T(n-6, 4) + T(n-7, 4) + T(n-9, 4), for n > 13.

%F T(n, 5) = T(n-1, 5) - T(n-3, 5) + T(n-5, 5) - T(n-6, 5) + T(n-8, 5) - T(n-9, 5) + T(n-11, 5) - T(n-13, 5) + T(n-14, 5), for n > 19.

%F T(n, 6) = T(n-2, 6) + T(n-3, 6) - 2*T(n-4, 6) - T(n-5, 6) + 2*T(n-7, 6) - 2*T(n-10, 6) + T(n-12, 6) + 2*T(n-13, 6) - T(n-14, 6) - T(n-15, 6) + T(n-17, 6), for n > 23.

%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,   1,   2,   1,   0,   1,   1;

%e   [8] 1,  -1,   0,   0,   2,  -1,   1,  -1,   1;

%e   [9] 1,   1,   0,  -1,  -2,   2,   0,   1,   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/(1-x+x^2). The coefficients of 1-x+x^2 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)=1+sum(k=1, n, squareRow(n-k, k+1)[k+1]*x^k)); Vec(1/f(x)+O(x^max_k)))

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

%Y Cf. A395191 (same idea but reversed order of terms relation in rows).

%K sign,easy,tabl

%O 0,32

%A _Thomas Scheuerle_, Apr 15 2026