A120113 Bi-diagonal inverse of number triangle A120101.

1, -6, 1, 0, -5, 1, 0, 0, -14, 1, 0, 0, 0, -3, 1, 0, 0, 0, 0, -11, 1, 0, 0, 0, 0, 0, -13, 1, 0, 0, 0, 0, 0, 0, -2, 1, 0, 0, 0, 0, 0, 0, 0, -17, 1, 0, 0, 0, 0, 0, 0, 0, 0, -19, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1

Offset

0,2

Comments

Subdiagonal is -A120114(n-1).

Links

G. C. Greubel, Rows n = 0..100 of the triangle, flattened

Formula

T(n, k) = 1 if k = n, T(n, k) = -A120114(n-1) if k = n-1, otherwise 0. - G. C. Greubel, May 05 2023

Example

Triangle begins
   1;
  -6,  1;
   0, -5,   1;
   0,  0, -14,  1;
   0,  0,   0, -3,   1;
   0,  0,   0,  0, -11,   1;
   0,  0,   0,  0,   0, -13,  1;
   0,  0,   0,  0,   0,   0, -2,   1;
   0,  0,   0,  0,   0,   0,  0, -17,   1;
   0,  0,   0,  0,   0,   0,  0,   0, -19,  1;
   0,  0,   0,  0,   0,   0,  0,   0,   0, -1,  1;

Mathematica

A120114[n_]:= LCM@@Range[2*n+4]/(LCM@@Range[2*n+2]);
A120113[n_, k_]:= If[k==n, 1, If[k==n-1, -A120114[n-1], 0]];
Table[A120113[n, k], {n, 0, 16}, {k, 0, n}]//Flatten

Prog

(Magma)
A120114:= func< n | Lcm([1..2*n+4])/Lcm([1..2*n+2]) >;
A120113:= func< n, k | k eq n select 1 else k eq n-1 select -A120114(n-1) else 0 >;
[A120113(n, k): k in [0..n], n in [0..16]]; // G. C. Greubel, May 05 2023
(SageMath)
def A120113(n, k):
    if (k<n-1): return 0
    elif (k==n-1): return -lcm(range(1, 2*n+3))/lcm(range(1, 2*n+1))
    else: return 1
flatten([[A120113(n, k) for k in range(n+1)] for n in range(17)]) # G. C. Greubel, May 05 2023

Crossrefs

Cf. A120101, A120111, A120114.

Sequence in context: A347237 A090203 A370563 * A074395 A355415 A262704

Adjacent sequences: A120110 A120111 A120112 * A120114 A120115 A120116

Keyword

sign,tabl

Author

Paul Barry, Jun 09 2006

Status

approved