A395191 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).

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, 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, 0, 0, -1, 1, 1, 1, 0, 0, 1, 7, 10, 3, 1, -1, 1, 1, 1, -1

Offset

0,31

Comments

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.

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

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.

Links

Table of n, a(n) for n=0..79.

Formula

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

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

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

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.

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.

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.

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.

Example

Triangle T(n, k) starts:
  [ 0] 1;
  [ 1] 1,  1;
  [ 2] 1, -1,  1;
  [ 3] 1,  1,  1,  1;
  [ 4] 1, -1,  0, -1,  1;
  [ 5] 1,  1,  0, -1,  1,  1;
  [ 6] 1, -1,  1,  0, -1, -1,  1;
  [ 7] 1,  1,  2,  2,  1,  0,  1,  1;
  [ 8] 1, -1,  2, -3,  2, -1,  1, -1,  1;
  [ 9] 1,  1,  1,  3,  4,  2,  0,  1,  1,  1;
  [10] 1, -1,  0, -2,  4, -7,  3,  0,  0, -1,  1;
  ...
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.

Prog

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

Crossrefs

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

Sequence in context: A380812 A016390 A327688 * A055800 A060572 A163543

Adjacent sequences: A395188 A395189 A395190 * A395192 A395193 A395194

Keyword

sign,easy,tabl

Author

Thomas Scheuerle, Apr 15 2026

Status

approved