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

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 first 15 terms: 1 2 4 6 16 12 64 24 36 48 1024 60 4096 192 144
 </pre>
+=={{header|BASIC}}==
+==={{header|PureBasic}}===
+{{trans|FreeBASIC}}
+<syntaxhighlight lang="purebasic">Procedure.i divisors(n)
+  ;find the number of divisors of an integer
+  Define.i r, i
+  r = 2
+  For i = 2 To n/2
+    If n % i = 0: r + 1
+    EndIf
+  Next i
+  ProcedureReturn r
+EndProcedure
+OpenConsole()
+Define.i UPTO, i, n, nfound
+UPTO = 15
+i = 2
+nfound = 1
+Dim sfound.i(UPTO)
+sfound(1) = 1
+While nfound < UPTO
+  n = divisors(i)
+  If n <= UPTO And sfound(n) = 0:
+    nfound + 1
+    sfound(n) = i
+  EndIf
+  i + 1
+Wend
+Print("1 ")    ;special case
+For i = 2 To UPTO
+  Print(Str(sfound(i)) + " ")
+Next i
+PrintN(#CRLF$ + "Press ENTER to exit"): Input()
+CloseConsole()</syntaxhighlight>
+==={{header|Yabasic}}===
+{{trans|FreeBASIC}}
+<syntaxhighlight lang="vbnet">UPTO = 15
+i = 2
+nfound = 1
+dim sfound(UPTO)
+sfound(1) = 1
+while nfound < UPTO
+    n = divisors(i)
+    if n <= UPTO and sfound(n) = 0 then
+        nfound = nfound + 1
+        sfound(n) = i
+    fi
+    i = i + 1
+end while
+print 1, " ";    //special case
+for i = 2 to UPTO
+    print sfound(i), " ";
+next i
+print
+end
+sub divisors(n)
+    local r, i
+    //find the number of divisors of an integer
+    r = 2
+    for i = 2 to n / 2
+        if mod(n, i) = 0  r = r + 1
+    next i
+    return r
+end sub</syntaxhighlight>
 =={{header|BCPL}}==