The Feynman point is a pattern in pi where the digit 9 repeats six times in a row. The 762nd to 767th digits are all 9. Here is pi to the Feynman point. Each row has 50 digits.
3.
14159265358979323846264338327950288419716939937510
58209749445923078164062862089986280348253421170679
82148086513282306647093844609550582231725359408128
48111745028410270193852110555964462294895493038196
44288109756659334461284756482337867831652712019091
45648566923460348610454326648213393607260249141273
72458700660631558817488152092096282925409171536436
78925903600113305305488204665213841469519415116094
33057270365759591953092186117381932611793105118548
07446237996274956735188575272489122793818301194912
98336733624406566430860213949463952247371907021798
60943702770539217176293176752384674818467669405132
00056812714526356082778577134275778960917363717872
14684409012249534301465495853710507922796892589235
42019956112129021960864034418159813629774771309960
51870721134999999...
1. Find three times where a digit repeats three times consecutively before the Feynman point.
2. The number 999 repeats twice consecutively in 999999. Find two 3-digit numbers that repeat twice consecutively before the Feynman point.
3. Find two 5-digit numbers that appear twice in different places before the Feynman point.
4. In one of the first 7 blocks of 100 digits, there is a 5-digit number ABCDE where ABCDE, ABCD, and BCDE all appear in different places in the same block. Which block of 100 digits is it, and what is ABCDE?
5. The number 999999 is of the form AABBAB, where A=B=9. Find a number of the form AABBAB before the Feynman point where A≠B.
6. Find a 6-digit number ABCDEF where ABCDEF and ABCDDEF both appear at different places before the Feynman point.
10 point 165
20 Pi=cutspc(str(#pi))
30 Blocks=mid(Pi,3,800)
40 print Blocks
50 Feynman=instr(Blocks,"999999"):print Feynman
55 print "Part 1:"
60 for Dig=0 to 9
70 S=cutspc(str(Dig)):S=S+S+S
75 Ix=0
80 repeat
90 Ix=instr(Ix+1,Blocks,S)
100 if Ix>0 and Ix<Feynman then print S,Ix
110 until Ix=0
120 next Dig
130 print:print "Part 2:"
140 for St=1 to Feynman-6
150 S=mid(Blocks,St,3)
160 if mid(Blocks,St+3,3)=S then print S+S,St
170 next St
180 print:print "Part 3:"
190 for St=1 to Feynman-5
200 S=mid(Blocks,St,5)
210 if instr(St+1,Blocks,S)>0 then print S,St,instr(St+1,Blocks,S)
220 next St
230 print:print "Part 4:"
240 for St=1 to 696
250 S5=mid(Blocks,St,5)
260 BlockNo=int((St-1)/100)+1
265 B=mid(Blocks,(BlockNo-1)*100+1,100)
270 Abcd=left(S5,4):Bcde=mid(S5,2,*)
275 St2=St-(BlockNo-1)*100
280 Tst=left(B,St2-1)+" "+mid(B,St2+5,*)
290 if instr(Tst,Abcd)>0 and instr(Tst,Bcde)>0 then print S5;St:print mid(Blocks,(BlockNo-1)*100+1,100)
300 next
310 print:print "Part 5:"
320 for St=1 to Feynman-6
330 if mid(Blocks,St,1)=mid(Blocks,St+1,1) and mid(Blocks,St,1)=mid(Blocks,St+4,1) then
340 :if mid(Blocks,St+2,1)=mid(Blocks,St+3,1) and mid(Blocks,St+2,1)=mid(Blocks,St+5,1) then
350 :if mid(Blocks,St+1,1)<>mid(Blocks,St+2,1) then print mid(Blocks,St,6),St
360 next
370 print:print "Part 6:"
380 for St=1 to Feynman-6
390 Abcddef=mid(Blocks,St,4)+mid(Blocks,St+3,3)
400 if instr(Blocks,Abcddef)>0 and instr(Blocks,Abcddef)<Feynman then print mid(Blocks,St,6),St,Abcddef,instr(Blocks,Abcddef)
410 next
finds
Part 1:
000 601
111 153
555 177
Part 2:
209209 326
305305 365
Part 3:
23846 16 579
60943 397 551
Part 4:
56482 225
4428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273
Part 5:
006606 307
Part 6:
502841 158 5028841 31
So,
In part 1,
000 begins at 601
111 begins at 153
555 begins at 177
Part 2 is similarly interpreted.
For part 3 both locations are listed for each string.
For part 4, string 56482 begins at 225. The block itself is shown below it, so you can see the 5648 and 6482 in their other locations.
Parts 5 and 6 show the respective locations of the shown strings of digits.
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Posted by Charlie
on 2012-02-29 15:35:48 |