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A342689
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Square array read by antidiagonals (upwards): A(n,k) = (k^Fibonacci(n) - 1) / (k - 1) for k >= 0 and n >= 0 with lim_{k -> 1} A(n,k) = A(n,1) = Fibonacci(n).
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0
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0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 1, 0, 1, 3, 3, 1, 1, 0, 1, 5, 7, 4, 1, 1, 0, 1, 8, 31, 13, 5, 1, 1, 0, 1, 13, 255, 121, 21, 6, 1, 1, 0, 1, 21, 8191, 3280, 341, 31, 7, 1, 1, 0, 1, 34, 2097151, 797161, 21845, 781, 43, 8, 1, 1, 0, 1, 55, 17179869184, 5230176601, 22369621, 97656, 1555, 57, 9, 1, 1, 0
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OFFSET
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0,12
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COMMENTS
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Replacing Fibonacci(n), A000045, with Lucas(n), A000032, you get another square array B(n,k). The terms satisfy the same recurrence equation B(n,k) = (k-1) * (B(n-1,k) * B(n-2,k) + B(n-1,k) + B(n-2,k) for k >= 0 and n > 1 with initial values B(0,k) = k+1 and B(1,k) = 1. Please take account of lim_{k -> 1} (k^Lucas(n) - 1) / (k - 1) = Lucas(n).
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LINKS
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FORMULA
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A(n,k) = (k - 1) * A(n-1,k) * A(n-2,k) + A(n-1,k) + A(n-2,k) for k >= 0 and n > 1 with initial values A(0,k) = 0 and A(1,k) = 1.
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EXAMPLE
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The array A(n,k) for k >= 0 and n >= 0 begins:
n \ k: 0 1 2 3 4 5 6 7 8 9 10 11
=========================================================================
0 : 0 0 0 0 0 0 0 0 0 0 0 0
1 : 1 1 1 1 1 1 1 1 1 1 1 1
2 : 1 1 1 1 1 1 1 1 1 1 1 1
3 : 1 2 3 4 5 6 7 8 9 10 11 12
4 : 1 3 7 13 21 31 43 57 73 91 111 133
5 : 1 5 31 121 341 781 1555 2801
6 : 1 8 255 3280 21845 97656
7 : 1 13 8191 797161 22369621
8 : 1 21 2097151 5230176601
9 : 1 34 17179869184
10 : 1 55
11 : 1 89
etc.
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CROSSREFS
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Cf. A011655 (column k = -1), A057427 (column 0), A000045 (column 1), A063896 (column 2), A000004 (row 0), A000012 (rows 1, 2), A000027 (row 3), A002061 (row 4), A053699 (row 5), A053717 (row 6), A060887 (row 7).
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KEYWORD
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AUTHOR
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STATUS
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approved
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