"...Though this be madness, yet there is method in 't.
12356 71011 12161 71823 24253 13233 40414 ..... etc
Find the method and provide the next 15 digits.
The first 3 integers then skip 1. 123 skip 4
The next 3 integers then skip 2. 567 skip 89
The next 3 integers then skip 3. 101112 skip 131415
The next 3 integers then skip 4. 161718 skip 19202122
... etc.
String them all together and put the digits into groups of five.
The first line below shows the first 50 digits... the original 35 and the next 15 asked for.
12356 71011 12161 71823 24253 13233 40414 25051 52616 26373
74758 68788 10010 11021 15116 11713 11321 33148 14915 01661
67168 18518 61872 05206 20722 62272 28248 24925 02712 72273
29529 62973 20321 32234 63473 48373 37437 54014 02403 43043
14324 60461 46249 14924 93523 52452 55565 57558 59059 15926
25626 62766 16626 63698 69970 07367 37738 77577 67778 15816
81785 68578 58898 89990
Table produced by:
DefDbl A-Z
Dim crlf$
Private Sub Form_Load()
Form1.Visible = True
Text1.Text = ""
crlf = Chr$(13) + Chr$(10)
prev = 0
For Skip = 1 To 40
For i = prev + 1 To prev + 3
s$ = s$ + LTrim(Str(i))
Next
prev = prev + Skip + 3
Next
Do
Text1.Text = Text1.Text & Left(s, 5) & " "
s = Mid(s, 6)
Loop Until Len(s) < 6
Text1.Text = Text1.Text & crlf & " done"
End Sub
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Posted by Charlie
on 2017-06-11 15:58:14 |
solution
