Abel Braaksma wrote:
Michael Kay wrote:
$fib[1] = 0 and $fib[2] = 1 and
every $i in 2 to count($fib) satisfies
$fib[$i] = $fib[$i - 1] + $fib[$i - 2]
Aha! (erlebnis)
For completeness, I will add a slightly corrected version (I saw the
light, thanks to Michael). The 'every' operator has lower precedence
than 'and', requiring parenthesis (otherwise you will receive an error)
and the '.. to ..' range should start with 3:
$fib[1] = 0 and $fib[2] = 1 and
(every $i in 3 to count($fib) satisfies
$fib[$i] = $fib[$i - 1] + $fib[$i - 2])
This returns true for example for the following Fibonacci sequence:
(0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597,
2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025)
Here's another version, which eases the expression a bit by removing the
special cases:
every $i in 1 to count($fib) satisfies
$fib[$i] = (0, $fib)[$i] + (0, 1, $fib)[$i]
-- Abel
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