A342149 Infinite square matrix A(m,n) = F(m+1) mod (n+1), m,n >= 1, where F = Fibonacci = A000045, read by falling antidiagonals.

1, 1, 0, 1, 2, 1, 1, 2, 0, 1, 1, 2, 3, 2, 0, 1, 2, 3, 1, 2, 1, 1, 2, 3, 0, 0, 1, 1, 1, 2, 3, 5, 3, 1, 0, 0, 1, 2, 3, 5, 2, 3, 1, 1, 1, 1, 2, 3, 5, 1, 1, 1, 2, 1, 1, 1, 2, 3, 5, 0, 6, 3, 4, 3, 2, 0, 1, 2, 3, 5, 8, 5, 0, 4, 0, 1, 0, 1, 1, 2, 3, 5, 8, 4, 5, 6, 1, 4, 0, 2, 1, 1, 2, 3, 5, 8, 3, 3, 2, 6, 5, 4, 1, 2, 0, 1, 2, 3, 5, 8, 2, 1, 7, 7, 5, 0, 3, 1, 1, 1

Offset

1,5

Comments

The determinant of the (upper left) n X n submatrix is zero iff n >= 35. (Observation by Bill Gosper, math-fun mailing list.)

The rows are eventually constant sequences, reaching the limit Fibonacci(m+1) after the final 0 in column n = Fibonacci(m+1); the sequence of columns converges to the Fibonacci sequence A000045 without the initial 0.

Links

Table of n, a(n) for n=1..120.

Bill Gosper, mysteriously vanishing sequence, math-fun mailing list, Jul 10 2021

Formula

A(m,n) = A342148(m+1,n+1) = A000045(m+1) mod (n+1).

See A342148 for more formulas.

Example

The matrix reads:
   [1  1  1  1  1  1  1 ...]
   [0  2  2  2  2  2  2 ...]
   [1  0  3  3  3  3  3 ...]
   [1  2  1  0  5  5  5 ...]
   [0  2  0  3  2  1  0 ...]
   [1  1  1  3  1  6  5 ...]
   [1  0  1  1  3  0  5 ...]
   (...)

Prog

(PARI) A342149(m, n)=fibonacci(m+1)%(n+1)
row(n)=[A342149(m, n-m+1) | m<-[1..n]] \\ The n-th falling antidiagonal.
concat([ row(n) | n <- [1..10] ]) \\ (beginning of) the "flattened" sequence.

Crossrefs

Cf. A000045, A342148 (matrix extended by an additional initial row and column).

Sequence in context: A157188 A329257 A173266 * A224326 A096496 A117209

Adjacent sequences: A342146 A342147 A342148 * A342150 A342151 A342152

Keyword

nonn,tabl

Author

M. F. Hasler, Jul 12 2021

Status

approved