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%I #10 Jul 13 2021 06:23:14
%S 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,
%T 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,
%U 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
%N Infinite square matrix A(m,n) = F(m+1) mod (n+1), m,n >= 1, where F = Fibonacci = A000045, read by falling antidiagonals.
%C The determinant of the (upper left) n X n submatrix is zero iff n >= 35. (Observation by _Bill Gosper_, math-fun mailing list.)
%C 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.
%H Bill Gosper, <a href="https://mailman.xmission.com/hyperkitty/list/math-fun@mailman.xmission.com/thread/TTKTZGDVDYS7UL5NZEMEEGHLY3YRWUMJ/">mysteriously vanishing sequence</a>, math-fun mailing list, Jul 10 2021
%F A(m,n) = A342148(m+1,n+1) = A000045(m+1) mod (n+1).
%F See A342148 for more formulas.
%e The matrix reads:
%e [1 1 1 1 1 1 1 ...]
%e [0 2 2 2 2 2 2 ...]
%e [1 0 3 3 3 3 3 ...]
%e [1 2 1 0 5 5 5 ...]
%e [0 2 0 3 2 1 0 ...]
%e [1 1 1 3 1 6 5 ...]
%e [1 0 1 1 3 0 5 ...]
%e (...)
%o (PARI) A342149(m,n)=fibonacci(m+1)%(n+1)
%o row(n)=[A342149(m,n-m+1) | m<-[1..n]] \\ The n-th falling antidiagonal.
%o concat([ row(n) | n <- [1..10] ]) \\ (beginning of) the "flattened" sequence.
%Y Cf. A000045, A342148 (matrix extended by an additional initial row and column).
%K nonn,tabl
%O 1,5
%A _M. F. Hasler_, Jul 12 2021
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