## Welcome to p196.org!

The following were some notes that I had written to myself for no other purpose than to remind myself to dig into it deeper at a later date. So far, I haven't done that...

7/14/03 Notes to self...

Sum the first and last digit of each iteration to get 1st column, sum second and last minus 1 of the same iteration to get 2nd column, sum third and last minus 2 of the same iteration to get 3rd column, etc...

So that:

3/5/8/ = 124...432 or 358...000 or 247...111

As a second example:

6/10/17/ = 258...954 or 199...815 or 338...973

This method of looking at the strings, without actually paying attention to the specific digits is credited to Vincent Prosper and Sebastian Vigenaue.

The digit sums will range from 1 to 18 for the first column, and 0-18 for the second, third, forth, etc. 0 being impossible, in the first column, since no iteration will sum a first and last 0. Although 0 is possible in any other column. 18 will be the maximum, since no sum will be greater than 9+9.

Then concatenate the iterations with a comma separating one iteration from the next. So that:

124...432,258...954 = 3/5/8/,6/10/17/

This provides us with an easily searchable string through a large number of iterations.

Let:

0 = a

1 = b

2 = c

3 = d

4 = e

5 = f

6 = g

7 = h

8 = i

9 = j

10 = k

11 = l

12 = m

13 = n

14 = p

15 = q

16 = r

17 = s

18 = t

The letter o is omitted to avoid the confusion with the number 0.

Now, any form that equals 3/5/8/,6/10/17/ can be written as dfi,gks. This might help eliminate the confusion of assuming that 3/5/8 are the actual digits of the iteration, when they are in fact the SUMS of multiple combinations of digits. For the remainder of this paper, when the letter notation is used, the "/" marks will be eliminated, since it is understood that each letter is a digit sum from the chart above.

This gives the following possible combinations for every digit sum:

t (18) - 1 combination 9/9

s (17) - 2 combinations 9/8 or 8/9

r (16) - 3 combinations 7/9 or 9/7 or 8/8

q (15) - 4 combinations 6/9 or 9/6 or 7/8 or 8/7

p (14) - 5 combinations 5/9 or 9/5 or 6/8 or 8/6 or 7/7

n (13) - 6 combinations 4/9 or 9/4 or 5/8 or 8/5 or 6/7 or 7/6

m (12) - 7 combinations 3/9 or 9/3 or 4/8 or 8/4 or 5/7 or 7/5 or 6/6

l (11) - 8 combinations 2/9 or 9/2 or 3/8 or 8/3 or 4/7 or 7/4 or 5/6 or 6/5

k (10) - 9 combinations 1/9 or 9/1 or 2/8 or 8/2 or 3/7 or 7/3 or 4/6 or 6/4 or 5/5

j (9) - 10 combinations 0/9 or 9/0 or 1/8 or 8/1 or 2/7 or 7/2 or 3/6 or 6/3 or 4/5 or 5/4

i (8) - 9 combinations 0/8 or 8/0 or 1/7 or 7/1 or 2/6 or 6/2 or 3/5 or 5/3 or 4/4

h (7) - 8 combinations 0/7 or 7/0 or 1/6 or 6/1 or 2/5 or 5/2 or 3/4 or 4/3

g (6) - 7 combinations 0/6 or 6/0 or 1/5 or 5/1 or 2/4 or 4/2 or 3/3

f (5) - 6 combinations 0/5 or 5/0 or 1/4 or 4/1 or 2/3 or 3/2

e (4) - 5 combinations 0/4 or 4/0 or 1/3 or 3/1 or 2/2

d (3) - 4 combinations 0/3 or 3/0 or 1/2 or 2/1

c (2) - 3 combinations 0/2 or 2/0 or 1/1

b (1) - 2 combination 0/1 or 1/0

a (0) - 1 combination 0/0

In the cases of "a" through "j", they can only be obtained in the first column from a 1/0, 2/0, 3/0 etc combination, and cannot be obtained by 0/1, 0/2 or 0/3 etc groupings, since 0 will never start an iteration string. For any remaining column, a 0/1, 0/2, 0/3 etc grouping is allowed and may occur.

The table above shows that the greatest number of possible combinations come from "j". Testing of the digit frequency within iterations has proven that digit pairs that sum 9, dominate the distribution. (DOES THIS GUARUNTEE THAT THE NEXT ITERATION WILL HAVE MORE CARRIES THAN THE CURRENT ONE?!?!? THAT THE “ENTROPY” IS INCREASED EVERY ITERATION??)

In an example iteration: "csa" it can be proven that no palindrome will form, since any letter "k" - "t" (Numbers 10 - 18) in the second column, will produce a carry in the next iteration, so "c" WILLNOT be of the form 1/1 in the next iteration. This would also apply to any first column of "e", "g", "i", "k", "m", "p", "r" or "t", where the second column is "k" through "t".

It is obvious that there are only two options for the addition of a string and it's reversal...

The first is that no digit sums will generate a carry. This would be a string that is composed solely of "a" through "j". This will result in a palindrome being formed. If the current iteration is an even number of digits, an even palindrome will result. If the current iteration is an odd number of digits, an odd palindrome will form.

The second option would be a string that will contain at least one carry in the addition. This may or may not form a palindrome. In most cases, no palindrome will be formed.

Jason Kruppa's work shows that if there is an odd number of digits in a string, the only way for the string to become a palindrome on the next iteration, is for the center "odd" digit to be a 0, and EVERY digit pair must sum 11 or 0. As a side note, these palindromes will an even number of digits.

If there is an EVEN number of digits in a string, the only way for the string to become a palindrome on the next iteration, every digit pair must sum 11 or 0. These palindromes will be an odd number of digits.

As examples, the strings 59026 or 443877 will become palindromes on the next iteration.

This work, involving carries in a string, forces the current iteration to be of the form "l...la", "la...a" or "l...l", in order for a palindrome to occur on the next iteration.

NOTE: It is CRITICALLY important to remember that the last letter in our psuedo-digit string, represents the MIDDLE most digit or pair of digits tested, NOT the end of the digit string.

In 127,986 iterations of the Lychrel number 196, there is only a SINGLE iteration of the pattern of "ll..." (11/11/x). It is "llb..." (11/11/1...) and occurs 51,909 iterations into the set. There is no second iteration that "matches" with this string. It occurs by itself as "11b", but nowhere does it occurs as "llb,xxx" except this single iteration. This seriously hinders the challenge of getting all of the digits to sum 0 or 11, for the formation of a palindrome!!

Since this is our only goal, something needs to account for the discrepancy between the digit sums dominated by 9 instead or 11.

24 iterations of 89 show:

89 s 17

187 it 8/16

968 sm 17/12

1837 il 8/11

9218 sd 17/3

17347 ilg 8/11/6

91718 scp 17/2/14

173437 ikh 8/10/7

907808 saq 17/0/15

1716517 iigm 8/8/6/12

8872688 rrne 16/16/13/4

17735476 hpli 7/14/11/8

85189247 qjds 15/9/3/17

159487405 gfnlr 6/5/13/11/16

664272356 mlhep 12/11/7/4/14

1317544822 dfjlj 3/5/9/11/9

3602001953 gljdba 6/11/9/3/1/0

7193004016 ncjha 13/2/9/7/0

13297007933 eglrha 4/6/11/16/7/0

47267087164 indnqa 8/13/3/13/15/0

93445163438 sgihlc 17/6/8/7/11/2

176881317877 ippqje 8/14/14/15/9/4

955594506548 sjkljj 17/9/10/11/9/9

1801200002107 iibdcaa 8/8/1/3/2/0/0

8813200023188 rrcgeaa 16/16/2/6/4/0/0

In this case, the iteration preceding the palindrome, will generate no carries, so a palindrome is formed. This is the only iteration of the set that does not contain a carried digit.

In the first 26 iterations of 196, every iteration produces a carried digit:

196 ht 7/18

887 nr 13/16

1675 gn 6/13

7436 nh 13/7

13783 elp 4/11/14

52514 jdk 9/3/10

94039 tha 18/7/0

187088 jrh 9/16/7

1067869 kgpp 10/6/14/14

10755470 bhlk 1/7/11/10

18211171 cqdc 2/15/3/2

35322452 fkhe 5/10/7/4

60744805 laqi 11/0/15/8

111589511 ccdpr 2/2/3/14/16

227574622 eenjp 4/4/13/9/14

454050344 ijhak 8/9/7/0/10

897100798 rtpba 16/18/14/1/0

1794102596 hrpgbb 7/16/14/6/1/1

8746117567 qnjncc 15/13/9/13/2/2

16403234045 gkeege 6/10/4/4/6/4

70446464506 najimi 13/0/9/8/12/8

130992928913 eejsll 4/4/9/17/11/11

450822227944 ijjqee 8/9/9/15/4/4

900544455998 sjjkjii 17/9/9/10/9/8/8

1800098901007 iiabatr 8/8/0/1/0/18/16

8801197801088 rracbsp 16/16/0/2/1/17/14

As an interesting side-note, looking at only the first digit from each iteration of the number 89, gives the string:

8 1 9 1 9 1 9 1 9 1 8 1 8 1 6 1 3 7 1 4 9 1 9 1 8 8 9

Looking at the first digit from the 196 iterations gives:

1 8 1 7 1 5 1 9 1 1 1 1 3 6 1 2 4 8 1 8 1 7 1 4 9 1 8

Lining the two up as best as possible:

8 1 9 1 9 1 9 1 9 1 8 1 8 1 6 1 3 7 1 4 9 1 9 1 8 8 9

1 8 1 7 1 5 1 9 1 1 1 1 3 6 1 2 4 8 1 8 1 7 1 4 9 1 8

Doesn't really show anything interesting, but if I use the longest string that repeats 100 times in 125,000 iterations the following shows up:

Iterations from 89: 8 1 9 1 9 1 9 1 9 1 8 1 8 1 6 1 3 7 1 4 9 1 9 1 8 8 9

Iterations from 196 (100 reps): 9 1 9 1 9 1 9 1 8 1 8 1 7 1 4 8 1 8 1 6 1 3 7 1 4 9 1 9 1 9 1

Iterations from 196 (1000 reps): 9 1 9 1 9 1 9 1 9 1 8 1 8 1

Iterations from 869 8 1 9 1 9 1 9 1 8 1 8 1 6 1 3 7 1 4 9 1 9 1 8

This is pretty much off topic for the rest of this paper, but it is kind of interesting to note that the same 81918191918181 pattern shows up in many different root numbers, even the ones that become a palindromes. I assume it is a function of the carried digit and it might also be related to the digit pairs summing 9 so much, but it's still odd to see...