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Frobenius numbers: Difference between revisions
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900591 919631* 937019 946719 958431 972179 986039
</pre>
=={{header|Nim}}==
As I like iterators, I used one for (odd) primes and one for Frobenius numbers. Of course, there are other ways to proceed.
<lang Nim>import sequtils, strutils
func isOddPrime(n: Positive): bool =
if n mod 3 == 0: return n == 3
var d = 5
while d * d <= n:
if n mod d == 0: return false
inc d, 2
if n mod d == 0: return false
inc d, 4
result = true
iterator oddPrimes(): int =
yield 3
var n = 5
while true:
if n.isOddPrime: yield n
inc n, 2
if n.isOddPrime: yield n
inc n, 4
iterator frobenius(lim: Positive): int =
var p1 = 2
for p2 in oddPrimes():
let f = p1 * p2 - p1 - p2
if f < lim: yield f
else: break
p1 = p2
const N = 10_000
var result = toSeq(frobenius(10_000))
echo "Found $1 Frobenius numbers less than $2:".format(result.len, N)
echo result.join(" ")</lang>
{{out}}
<pre>Found 25 Frobenius numbers less than 10000:
1 7 23 59 119 191 287 395 615 839 1079 1439 1679 1931 2391 3015 3479 3959 4619 5039 5615 6395 7215 8447 9599</pre>
=={{header|Perl}}==
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