Stern-Brocot sequence: Difference between revisions
added Arturo
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Drkameleon (talk | contribs) (added Arturo) |
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true</pre>
=={{header|Arturo}}==
<syntaxhighlight lang="arturo">sternBrocot: function [mx][
seq: [1 1]
result: [[1 1] [2 1]]
idx: 1
while [idx < mx][
'seq ++ seq\[idx] + seq\[idx - 1]
'result ++ @[@[size seq, last seq]]
'seq ++ seq\[idx]
'result ++ @[@[size seq, last seq]]
inc 'idx
]
return result
]
sbs: sternBrocot 1000
print ["First 15 terms:" join.with:", " first.n:15 map sbs 'sb -> to :string last sb]
print ""
indexes: array.of:101 0
toFind: 101
loop sbs 'sb [
[i, n]: sb
if and? -> contains? 1..100 n -> zero? indexes\[n][
indexes\[n]: i
dec 'toFind
if zero? toFind [
break
]
]
]
loop (@1..10) ++ 100 'n ->
print ["Index of first occurrence of number" n ":" indexes\[n]]
print ""
prev: 1
idx: 1
loop sbs 'sb [
[i, n]: sb
if not? one? gcd @[prev n] -> break
prev: n
inc 'idx
if idx > 1000 -> break
]
print (idx =< 1000)? -> ["Found two successive terms at index:" idx]
-> "All consecutive terms up to the 1000th member have a GCD equal to one."</syntaxhighlight>
{{out}}
<pre>First 15 terms: 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4
Index of first occurrence of number 1 : 1
Index of first occurrence of number 2 : 3
Index of first occurrence of number 3 : 5
Index of first occurrence of number 4 : 9
Index of first occurrence of number 5 : 11
Index of first occurrence of number 6 : 33
Index of first occurrence of number 7 : 19
Index of first occurrence of number 8 : 21
Index of first occurrence of number 9 : 35
Index of first occurrence of number 10 : 39
Index of first occurrence of number 100 : 1179
All consecutive terms up to the 1000th member have a GCD equal to one.</pre>
=={{header|AutoHotkey}}==
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