Sequence: smallest number with exactly n divisors: Difference between revisions

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 =={{header|Phix}}==
-Using the formulae from the OEIS:A005179 link above.<br>
+Using the various formula from the OEIS:A005179 link above.<br>
-Uses primes[] and add_block() from [[Extensible_prime_generator#Phix|Extensible_prime_generator]],
+get_primes() and product() have recently been added as new builtins, if necessary see [[Extensible_prime_generator#Phix|Extensible_prime_generator]] and [[Deconvolution/2D%2B#Phix]].
-product() has recently been added as a new builtin, if you need it see [[Deconvolution/2D%2B#Phix]].
 <lang Phix>constant limit = iff(machine_bits()=32?58:66)
 sequence found = repeat(0,limit)
 integer n = 1
-atom ri
+procedure populate_found(integer i)
+    while found[i]=0 do
+        integer k = length(factors(n,1))
+        if k<=limit and found[k]=0 then
+            found[k] = n
+        end if
+        n += 1
+    end while
+end procedure
 for i=1 to limit do
     sequence f = factors(i,1)
     integer lf = length(f)
-    if lf<=2 then
+    atom ri
-        ri = power(2,i-1)                               -- prime (or 1)
+    if lf<=2 then                       ri = power(2,i-1)                               -- prime (or 1)
+    elsif lf=3 then                     ri = power(6,f[2]-1)                            -- p^2 (eg f={1,5,25})
-    elsif lf=3 then
-        ri = power(6,f[2]-1)                            -- p^2 (eg f={1,5,25})
     elsif f[2]>2 -- (see note)
-      and f[$] = power(f[2],lf-1) then
+      and f[$] = power(f[2],lf-1) then  ri = power(product(get_primes(-(lf-1))),f[2]-1) -- p^k (eg f={1,3,9,27})
+    elsif length(f)=4 then              ri = power(2,f[3]-1)*power(3,f[2]-1)            -- p*q (eg f={1,2,3,6})
-        while length(primes)<lf-1 do
+    else populate_found(i)              ri = found[i]                                   -- do the rest manually
-            add_block()
-        end while
-        ri = power(product(primes[1..lf-1]),f[2]-1)     -- p^k (eg f={1,3,9,27})
-    elsif length(f)=4 then
-        ri = power(2,f[3]-1)*power(3,f[2]-1)            -- p*q (eg f={1,2,3,6})
-    else
-        while found[i]=0 do
-            integer k = length(factors(n,1))            -- do the rest manually
-            if k<=limit and found[k]=0 then
-                found[k] = n
-            end if
-            n += 1
-        end while
-        ri = found[i]
     end if
     printf(1,"%d->%d\n",{i,ri})
 end for</lang>
-note: the f[2]>2 test should really be something more like >log(primes[lf-1])/log(2),
+Note: the f[2]>2 test should really be something more like >log(primes[lf-1])/log(2),
-apparently, but everything seems ok within the IEE754 53/64 bit limits this imposes.
+apparently, but everything seems ok within the IEEE 754 53/64 bit limits this imposes.
-afaict, it takes longer to print the answers that it did to calculate them, tee hee!
+It takes longer, afaict, to print the answers that it did to calculate them, tee hee!
 {{out}}
 -bit (as shown) manages 8 more answers than 32-bit, which as per limit halts on 58.