There is a finite number of
Ulam numbers n, such that n+1 is also an Ulam number.
List all such pairs.
The program below ran overnight and got up to n=5843 (that is, identified any Ulam numbers that were under that value, not that many Ulam numbers) before I stopped it. It started only at the 2 to 3 level and so excluded the 1. I manually added the 1, 2 pair to the output below, making 4 pairs altogether. Obviously not a proof of completeness.
ordinal Ulam
Ulam # Number
itself
1 1
2 2
2 2
3 3
3 3
4 4
15 47
16 48
Sloane's OEIS has a note that
1,2,3,47 are the only values of x < 6.759*10^8 such that x and x+1 are both Ulam numbers. - Jud McCranie, Jun 08 2001. This holds through the first 28 billion Ulam numbers - Jud McCranie, Jan 07 2016.
Jud must have a faster computer or a better algorithm (inclusive OR). But still doesn't look like a proof of completeness (proof of finite nature of the set) had been achieved.
DefDbl A-Z
Dim crlf$, u(1000), ways, sum(1000), ptr, term(2), terms, tot, n, t1, t2
Private Sub Form_Load()
Form1.Visible = True
Text1.Text = ""
crlf = Chr$(13) + Chr$(10)
u(1) = 1: u(2) = 2: ptr = 3
sum(1) = 1: sum(2) = 3
For n = 3 To 10000
DoEvents
ways = 0: tot = 0: terms = 0
addon ptr - 1
If ways = 1 Then
u(ptr) = n
sum(ptr) = sum(ptr - 1) + n
If u(ptr - 1) = n - 1 Then
Text1.Text = Text1.Text & ptr - 1 & Str(u(ptr - 1)) & crlf
Text1.Text = Text1.Text & ptr & Str(n) & crlf
Text1.Text = Text1.Text & crlf
End If
ptr = ptr + 1
End If
Next
Text1.Text = Text1.Text & crlf & " done"
End Sub
Sub addon(wh)
If u(wh) <= n - tot And sum(wh) >= n - tot And terms < 2 Then
terms = terms + 1
term(terms) = wh
tot = tot + u(wh)
If tot = n Then
ways = ways + 1
t1 = term(1): t2 = term(2)
Else
addon (wh - 1)
End If
tot = tot - u(wh)
term(terms) = 0
terms = terms - 1
End If
w = wh - 1
If w > 0 Then
If sum(w) >= n - tot Then
addon w
End If
End If
End Sub
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Posted by Charlie
on 2017-11-18 16:10:30 |
computer "solution"