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An interesting recurrence (Posted on 2017-01-17) Difficulty: 3 of 5
A sequence with the recurrence f(n)=3*f(n-1)+f(n-2) starts with two 1-digit numbers. The sequence contains the 8-digit number ABCDAECD. A≠0, and A, B, C, D, and E are not necessarily distinct. Find all possible values of ABCDAECD.

No Solution Yet Submitted by Math Man    
Rating: 5.0000 (1 votes)

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computer solution | Comment 2 of 5 |
With A, B, C, D and E not necessarily being distinct:

terms
1 2   ABCDAECD

0 3   15291729
0 3   50505150
0 6   30583458
1 1   21932293
1 1   72437443
1 4   21932293
1 4   72437443
2 2   13281128
2 2   43864586
2 5   28572857
2 5   94369736
2 8   13281128
2 8   43864586
3 0   15291729
3 0   50505150
3 3   19921692
3 3   65796879
3 6   10661766
3 6   35213421
3 9   15291729
3 9   50505150
4 4   26562256
5 2   17911091
5 8   48494549
6 0   30583458
6 3   81088608
6 6   39843384
6 9   55135113
7 1   52515751
9 9   59765076

There are 30 rows above but there are only 21 distinct values for ABCDAECD:

10661766
13281128
15291729
17911091
19921692
21932293
26562256
28572857
30583458
35213421
39843384
43864586
48494549
50505150
52515751
55135113
59765076
65796879
72437443
81088608
94369736

DefDbl A-Z
Dim crlf$


Private Sub Form_Load()
 Form1.Visible = True
 
 
 Text1.Text = ""
 crlf = Chr$(13) + Chr$(10)
 
 For a = 0 To 9
 For b = 0 To 9
   x = a: y = b
   While x < 100000000 And (x > 0 Or y > 0)
    DoEvents
      z = 3 * y + x
      zs$ = LTrim(Str(z))
      If Len(zs) = 8 Then
      If Mid(zs, 1, 1) = Mid(zs, 5, 1) And Mid(zs, 3, 1) = Mid(zs, 7, 1) And Mid(zs, 4, 1) = Mid(zs, 8, 1) Then
      '  If Mid(zs, 1, 1) <> Mid(zs, 2, 1) And Mid(zs, 1, 1) <> Mid(zs, 3, 1) And Mid(zs, 1, 1) <> Mid(zs, 4, 1) Then
      '  If Mid(zs, 2, 1) <> Mid(zs, 3, 1) And Mid(zs, 2, 1) <> Mid(zs, 4, 1) And Mid(zs, 2, 1) <> Mid(zs, 6, 1) And Mid(zs, 2, 1) <> Mid(zs, 6, 1)  Then
      '    If Mid(zs, 3, 1) <> Mid(zs, 4, 1) And Mid(zs, 3, 1) <> Mid(zs, 6, 1)  Then
           Text1.Text = Text1.Text & a & Str(b) & "   " & z & crlf
      '    End If
      '  End If
      '  End If
      End If
      End If
      x = y: y = z
   Wend
 Next b
 Next a
 
  
 Text1.Text = Text1.Text & crlf & " done"
  
End Sub

Notice the commented out code would make A, B, C, D and E necessarily distinct. With that code restored, to require distinct A, B, C, D and E, the results would be:

0 3   15291729
0 6   30583458
2 2   43864586
2 5   94369736
2 8   43864586
3 0   15291729
3 3   65796879
3 6   35213421
3 9   15291729
6 0   30583458
9 9   59765076

These are 7 distinct values:

15291729
30583458
35213421
43864586
59765076
65796879
94369736



Note that in many cases, a given starting pair, such as 1,4, will produce two different numbers in the same sequence that fit the pattern.

Edited on January 17, 2017, 11:34 am
  Posted by Charlie on 2017-01-17 11:20:42

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