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A342149
Infinite square matrix A(m,n) = F(m+1) mod (n+1), m,n >= 1, where F = Fibonacci = A000045, read by falling antidiagonals.
2
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
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
KEYWORD
nonn,tabl
AUTHOR
M. F. Hasler, Jul 12 2021
STATUS
approved