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 A chain of powers (Posted on 2016-12-22)
Let f1(n)=(nn)mod7
Let f2(n)=((f1(n))n)mod7
Let f3(n)=((f2(n))n)mod7
etc

Evaluate f7(n) for n=1,2....9

 No Solution Yet Submitted by Ady TZIDON No Rating

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 computer observations Comment 1 of 1
` n    f1(n) thru f7(n)`
` 1      1 1 1 1 1 1 1 2      4 2 4 2 4 2 4 3      6 6 6 6 6 6 6 4      4 4 4 4 4 4 4 5      3 5 3 5 3 5 3 6      1 1 1 1 1 1 1 7      0 0 0 0 0 0 0 8      1 1 1 1 1 1 1 9      1 1 1 1 1 1 110      4 4 4 4 4 4 411      2 4 2 4 2 4 212      1 1 1 1 1 1 113      6 6 6 6 6 6 614      0 0 0 0 0 0 015      1 1 1 1 1 1 116      2 2 2 2 2 2 217      5 3 5 3 5 3 518      1 1 1 1 1 1 119      5 5 5 5 5 5 520      1 1 1 1 1 1 121      0 0 0 0 0 0 022      1 1 1 1 1 1 123      4 2 4 2 4 2 424      1 1 1 1 1 1 125      4 4 4 4 4 4 426      4 2 4 2 4 2 427      6 6 6 6 6 6 628      0 0 0 0 0 0 029      1 1 1 1 1 1 130      1 1 1 1 1 1 1`

I don't see a pattern to f7(n). Within a given n, the cycle length is only 1 or 2 going from fx(n) to f(x+1)(n). As a result f7 matches f1, which is n^n mod 7.

Form1.Visible = True

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

For n = 1 To 30
f1 = 1
For i = 1 To n
f1 = f1 * n Mod 7
Next
f2 = 1
For i = 1 To n
f2 = f2 * f1 Mod 7
Next
f3 = 1
For i = 1 To n
f3 = f3 * f2 Mod 7
Next
f4 = 1
For i = 1 To n
f4 = f4 * f3 Mod 7
Next
f5 = 1
For i = 1 To n
f5 = f5 * f4 Mod 7
Next
f6 = 1
For i = 1 To n
f6 = f6 * f5 Mod 7
Next
f7 = 1
For i = 1 To n
f7 = f7 * f6 Mod 7
Next
Text1.Text = Text1.Text & mform(n, "#0") & "     " & Str(f1) & Str(f2) & Str(f3) & Str(f4) & Str(f5) & Str(f6) & Str(f7) & crlf
Next

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

End Sub

Function mform\$(x, t\$)
a\$ = Format\$(x, t\$)
If Len(a\$) < Len(t\$) Then a\$ = Space\$(Len(t\$) - Len(a\$)) & a\$
mform\$ = a\$
End Function

 Posted by Charlie on 2016-12-22 09:37:15

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