A144083 Triangle read by rows: partial sums from the right of an A010892 subsequences decrescendo triangle.

1, 2, 1, 2, 2, 1, 1, 2, 2, 1, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 1, 0, 0, 1, 2, 2, 1, 2, 1, 0, 0, 1, 2, 2, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1

Offset

0,2

Comments

n-th row = (n+1) terms of an infinitely periodic cycle: (..., 1, 0, 0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1), shifting to the right one place for the next row.

Construct an A010892 decrescendo triangle: (1; 1,1; 0,1,1; -1,0,1,1; ...) and take partial sums starting from the right.

Links

Table of n, a(n) for n=0..77.

Formula

T(n, k) = 1 + A010892(n-k-1), with 0 <= k <= n. - Stefano Spezia, Feb 11 2023

Example

First few rows of the triangle:
  1;
  2, 1;
  2, 2, 1;
  1, 2, 2, 1;
  0, 1, 2, 2, 1;
  0, 0, 1, 2, 2, 1;
  1, 0, 0, 1, 2, 2, 1;
  2, 1, 0, 0, 1, 2, 2, 1;
  2, 2, 1, 0, 0, 1, 2, 2, 1;
  1, 2, 2, 1, 0, 0, 1, 2, 2, 1;
  0, 1, 2, 2, 1, 0, 0, 1, 2, 2, 1;
  ...
Row 3 = (1, 2, 2, 1) = partial sums of (-1, 0, 1, 1).

Mathematica

A010892[n_]:={1, 1, 0, -1, -1, 0}[[Mod[n, 6]+1]]; T[n_, k_]:=1+A010892[n-k-1]; Table[T[n, k], {n, 0, 11}, {k, 0, n}]//Flatten (* Stefano Spezia, Feb 11 2023 *)

Crossrefs

Cf. A010892, A077859 (row sums), A164965 (1st column).

Sequence in context: A156257 A097867 A075344 * A332292 A054350 A026606

Adjacent sequences: A144080 A144081 A144082 * A144084 A144085 A144086

Keyword

nonn,tabl

Author

Gary W. Adamson, Sep 10 2008

Status

approved