All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Odd Egyptians (Posted on 2017-06-21)
Find all ways of expressing 1 as the sum of nine distinct odd Egyptian Fractions.

I.e. reciprocals of odd whole numbers.

 No Solution Yet Submitted by Jer No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Programming exercise (spoiler); 20/20 hindsight | Comment 1 of 2
The program finds

1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/15 + 1/21 + 1/135 + 1/10395

1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/15 + 1/21 + 1/165 + 1/693

1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/15 + 1/21 + 1/231 + 1/315

1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/15 + 1/33 + 1/45 + 1/385

1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/15 + 1/35 + 1/45 + 1/231

The whole output was

2.828968253968251
+ 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/15 + 1/21 + 1/135 + 1/10395
1
+ 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/15 + 1/21 + 1/165 + 1/693
1
+ 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/15 + 1/21 + 1/231 + 1/315
1
+ 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/15 + 1/33 + 1/45 + 1/385
1
+ 1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/15 + 1/35 + 1/45 + 1/231

19999

done

The first number printed was just a verification that the reference table was constructed correctly, by checking its highest entry. The table was used in the program to see if there were sufficiently large numbers in the next few to get to at least 1 as a total. Combined with stopping the recursion if the total got to over 1 or the number of terms would get to more than 9, this speeded up processing.

The number 19999 at the end indicated we reached the limit of the entries we were trying for. If it had stopped short of that value, we could say we reached a natural limit and that there were no further answers. But that's not the case so we haven't show that we have all the answers. This was probably to be expected as the series diverges, so this method would never show a lack of possibility for further solutions.

Discussions while in the queue, with reference to outside web sites, indicate these are the only five possibilities, but this program does not prove that. The very determination to go to 20,000 was based on a knowledge of the answer, and so is really a cheat.

DefDbl A-Z
Dim crlf\$, h(36), tot, sumNext(1 To 9, 20000), highI

Form1.Visible = True

Text1.Text = ""
crlf = Chr\$(13) + Chr\$(10)

For i = 20000 To 1 Step -1
sumNext(1, i) = 1 / i
For n = 2 To 9
For j = i To i + n - 1
If j <= 20000 Then
sumNext(n, i) = sumNext(n, i) + sumNext(1, j)
End If
Next
Next
Next

Text1.Text = Text1.Text & sumNext(9, 1)

For st = 1 To 101 Step 2
h(1) = st: tot = 1 / st
Next

Text1.Text = Text1.Text & crlf & highI & crlf

Text1.Text = Text1.Text & crlf & " done"

End Sub

DoEvents
st = h(wh - 1) + 2
For i = st To 20000 Step 2
DoEvents
h(wh) = i
savetot = tot
tot = tot + sumNext(1, i)
If Abs(tot - 1) < 0.0000000000001 And wh = 9 Then
Text1.Text = Text1.Text & tot & crlf & "   "
For j = 1 To wh
Text1.Text = Text1.Text & " + 1/" & h(j)
Next
Text1.Text = Text1.Text & crlf
End If
If i < 20000 And tot < 1.0001 And wh < 9 Then
If tot + sumNext(9 - wh, i + 1) + 0.000000001 >= 1 Then
If i > highI Then highI = i
End If
End If
tot = savetot
Next
End Sub

 Posted by Charlie on 2017-06-21 16:02:27

 Search: Search body:
Forums (0)