Sequence: nth number with exactly n divisors: Difference between revisions
Sequence: nth number with exactly n divisors (view source)
Revision as of 02:01, 25 May 2021
, 3 years ago→{{header|Sidef}}
Line 1,686:
done...
</pre>
=={{header|Ruby}}==
{{trans|Java}}
<lang ruby>def isPrime(n)
return false if n < 2
return n == 2 if n % 2 == 0
return n == 3 if n % 3 == 0
k = 5
while k * k <= n
return false if n % k == 0
k = k + 2
end
return true
end
def getSmallPrimes(numPrimes)
smallPrimes = [2]
count = 0
n = 3
while count < numPrimes
if isPrime(n) then
smallPrimes << n
count = count + 1
end
n = n + 2
end
return smallPrimes
end
def getDivisorCount(n)
count = 1
while n % 2 == 0
n = (n / 2).floor
count = count + 1
end
d = 3
while d * d <= n
q = (n / d).floor
r = n % d
dc = 0
while r == 0
dc = dc + count
n = q
q = (n / d).floor
r = n % d
end
count = count + dc
d = d + 2
end
if n != 1 then
count = 2 * count
end
return count
end
MAX = 15
@smallPrimes = getSmallPrimes(MAX)
def OEISA073916(n)
if isPrime(n) then
return @smallPrimes[n - 1] ** (n - 1)
end
count = 0
result = 0
i = 1
while count < n
if n % 2 == 1 then
# The solution for an odd (non-prime) term is always a square number
root = Math.sqrt(i)
if root * root != i then
i = i + 1
next
end
end
if getDivisorCount(i) == n then
count = count + 1
result = i
end
i = i + 1
end
return result
end
n = 1
while n <= MAX
print "A073916(", n, ") = ", OEISA073916(n), "\n"
n = n + 1
end</lang>
{{out}}
<pre>A073916(1) = 1
A073916(2) = 3
A073916(3) = 25
A073916(4) = 14
A073916(5) = 14641
A073916(6) = 44
A073916(7) = 24137569
A073916(8) = 70
A073916(9) = 1089
A073916(10) = 405
A073916(11) = 819628286980801
A073916(12) = 160
A073916(13) = 22563490300366186081
A073916(14) = 2752
A073916(15) = 9801</pre>
=={{header|Sidef}}==
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