Erdős–Woods numbers

Description

A positive integer k is said to be an Erdős–Woods number if it has the following property: there exists a positive integer a such that in the sequence (a, a + 1, …, a + k) of consecutive integers, each of the elements has a non-trivial common factor with one of the endpoints. In other words, k is an Erdős–Woods number if there exists a positive integer a such that for each integer i between 0 and k, at least one of the greatest common divisors gcd(a, a + i) or gcd(a + i, a + k) is greater than 1.

It can be shown that there are infinitely many such numbers. Moreover, if k is an E-W number for some integer a, then one can find an infinite number of other a's using the formula a(jq + 1) where:

q is the product of all odd prime factors of a + k; and
j is any positive integer.

Example

16 is an E-W number (and also the first) for which the smallest value of a is 2184. This is because in the sequence 2184 to 2200 inclusive, the prime factors of the endpoints are:

2³ x 3 x 7 x 13 = 2184
2³ x 5² x 11 = 2200

and, if you check all the numbers between them, you will find that they all have a prime factor in common with at least one of the endpoints (2189 is divisible by 11, 2191 by 7, 2197 by 13 and the rest by either 2, 3, or 5).

Task

Compute and show here the first 20 E-W numbers together with their corresponding smallest value of a. If your language doesn't support arbitrary precision arithmetic, just compute and show as many as you can.

Extra credit

Do the same for the next 20 E-W numbers.

Note

Although all the E-W numbers relevant to this task are even, odd examples do exist the first one being k = 903.

References

Wren

Library: Wren-big

Library: Wren-fmt

Library: Wren-sort

Library: Wren-trait

It's not easy to find a way of doing this in a reasonable time.

I ended up translating the Python 3.8 code (the more readable version) here, 6th post down, and found the first 20 E-W numbers in around 93 seconds. Much slower than Python which has arbitrary precision numerics built-in nowadays but acceptable for Wren. <lang ecmascript>import "./big" for BigInt import "./fmt" for Conv, Fmt import "./sort" for Sort import "./trait" for Indexed

var zero = BigInt.zero var one = BigInt.one var two = BigInt.two

var ew = Fn.new { |n|

   var primes = []
   var k = 1
   var P = one
   while (k < n) {
       if ((P % k) != 0) primes.add(k)
       P = P * k * k
       k = k + 1
   }
   var divs = []
   var np = primes.count
   if (np > 0) {
       for (a in 0...n) {
           var A = BigInt.new(a)
           var s = primes.map { |p| Conv.btoi(A % p == 0).toString }.join()[-1..0]
           divs.add(Conv.atoi(s, 2))
       }
   }
   var partitions = [ [zero, zero, two.pow(np) - one] ]
   var key = Fn.new { |x| Conv.bin(divs[x] | divs[n-x])[-1..0].indexOf("1") }
   var cmp = Fn.new { |i, j| (key.call(j) - key.call(i)).sign }
   for (i in Sort.merge((1...n).toList, cmp)) {
       var newPartitions = []
       var factors = BigInt.new(divs[i])
       var otherFactors = BigInt.new(divs[n-i])
       for (p in partitions) {
           var setA = p[0]
           var setB = p[1]
           var rPrimes = p[2]
           if ((factors & setA) != zero || (otherFactors & setB) != zero) {
               newPartitions.add(p)
               continue
           }
           for (se in Indexed.new((factors & rPrimes).toBaseString(2)[-1..0])) {
               var ix = se.index
               var v = se.value
               if (v == "1") {
                   var w = one << ix
                   newPartitions.add([setA ^ w, setB, rPrimes ^ w])
               }
           }
           for (se in Indexed.new((otherFactors & rPrimes).toBaseString(2)[-1..0])) {
               var ix = se.index
               var v = se.value
               if (v == "1") {
                   var w = one << ix
                   newPartitions.add([setA, setB ^ w, rPrimes ^ w])
               }
           }
       }
       partitions = newPartitions
   }
   var result = null
   for (p in partitions) {
       var px = p[0]
       var py = p[1]
       var x = one
       var y = one
       for (p in primes) {
           if ((px % two) == one) x = x * p
           if ((py % two) == one) y = y * p
           px = px / two
           py = py / two
       }
       var N = BigInt.new(n)
       var temp = x.modPow(-1, y) * N % y * x - N
       result = result ? BigInt.min(result, temp) : temp
   }
   return result

}

var k = 3 var count = 0 System.print("THe first 20 Erdős–Woods numbers and their minimum interval start values are:") while (count < 20) {

   var a = ew.call(k)
   if (a) {
       Fmt.print("$3d -> $i", k, a)
       count = count + 1
   }
   k = k + 1

}</lang>

Output:

THe first 20 Erdős–Woods numbers and their minimum interval start values are:
 16 -> 2184
 22 -> 3521210
 34 -> 47563752566
 36 -> 12913165320
 46 -> 21653939146794
 56 -> 172481165966593120
 64 -> 808852298577787631376
 66 -> 91307018384081053554
 70 -> 1172783000213391981960
 76 -> 26214699169906862478864
 78 -> 27070317575988954996883440
 86 -> 92274830076590427944007586984
 88 -> 3061406404565905778785058155412
 92 -> 549490357654372954691289040
 94 -> 38646299993451631575358983576
 96 -> 50130345826827726114787486830
100 -> 35631233179526020414978681410
106 -> 200414275126007376521127533663324
112 -> 1022681262163316216977769066573892020
116 -> 199354011780827861571272685278371171794