From: ksbrown@ksbrown.seanet.com (Kevin Brown)
Newsgroups: can.schoolnet.math.jr,can.schoolnet.math.sr,sci.math
Subject: Re: Palindromic Numbers
Date: Mon, 19 Feb 1996 18:39:50 GMT
Bob Shingleton wrote:
> The number 17 is not palindromic, but if its digits are reversed to
> give 71, then the sum 17+71=88 which is a palindromic number. The
> number 19 needs 2 reversals and additions to become a Palindromic
> number 19+91=110, 110+011=121. How many reversals does 89 need?
rmorewoo@cln.etc.bc.ca (Robert Morewood) wrote:
> I get 24 steps ending with the palindrome: 8813200023188.
> ...I followed 196 for 7500 flips (let the computer run all night)
> and still no palindrome. Has anyone any idea of how to show it
> NEVER gives a palindrome?
There's no known proof that any sequence of "digit reversal sums" is
palindrome-free when working in base 10. However, it isn't hard to
prove the existence of sequences that never produce a palindrome in
certain other bases. For example, the smallest number that never
becomes palindromic in the base 2 is 10110 (decimal 22). To prove
this, first observe that the reverse-sum sequence beginning with
10110 is
10110
100011
1010100
1101001
10110100
etc.
The last term quoted above is 10110100, which is of the form
10 [n*1] 01 [n*0]
where the symbol [n*x] signifies n consecutive occurences of the digit
x. By simple arithmetic we can demonstrate that the reverse-sum
sequence beginning with any number of this form proceedes as follows
10 [n*1] 01 [n*0]
11 [(n-2)*0] 1000 [(n-2)*1] 01
10 [n*1] 01 [(n+1)*0]
11 [n*0] 10 [(n-1)*1] 01
10 [(n+1)*1] 01 [(n+1)*0]
The last representation is identical to the first, except that n
has been replaced by n+1. By induction, it follows that the entire
sequence consists of repetitions of this cycle, and none of the
elements are palindromes.
In the base 4, the number 255 (decimal) leads to a palindrome-free
sequence with the following six-step cycle
22 [n*0] 131 [n*3] 12
10 [(n+2)*3] 23 [(n+2)*0]
11 [n*0] 3222 [n*3] 01
22 [n*0] 2111 [n*3] 12
10 [(n+2)*3] 23 [(n+3)*0]
11 [(n+1)*0] 312 [(n+1)*3] 01
22 [(n+1)*0] 131 [(n+1)*3] 12
If a similar sort of argument can be constructed for palindrome-free
sequences in the base 10, it would probably be much more complicated,
and based on a much longer formal cycle.
The smallest numbers leading to palindrome-free sequences in each base
from 2 through 19 are listed below (in decimal):
2 22 8 1021 14 361
3 100 9 593 15 447
4 255 10 196 16 413
5 708 11 1011 17 3297
6 1079 12 237 18 519
7 2656 13 2196 19 341
It's interesting that, in each base, all the palindrome-free sequences
converge very rapidly on just a small number of sequences. For
example, there are 63 decimal numbers less than or equal to 4619 that
[evidently] never become palindromic, and these 63 numbers each lead
to one of only three palindrome-free sequences. The initial values of
these sequences are
A B C
------ ------- --------
887 1857 9988
1675 9438 18887
7436 17787 97768
13783 96558 184547
52514 182127 930028
94039 903408 1750067
187088 1707717 9350638
1067869 8884788 17711177
etc. etc. etc.
I suspect these sequence are cyclical (in the sense of the base
2 and base 4 cycles described above), but with irrational periods.
Notice that each term in the sequence can be regarded as a sort of
"convolution" of the preceeding term, and there are known examples
of sequences based on convolution that are cyclical with irrational
periods.
=====================================================================
MathPages at --> http://www.seanet.com/~ksbrown/
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From: stl137@aol.com (STL137)
Newsgroups: sci.math
Subject: Re: numbers, mirror numbers
Date: 20 Oct 1998 01:17:13 GMT
In binary and hex, it has been proven that some numbers go forever. 196 is an
open question in the decimal system. See "Two years of computing" at
http://www.fourmilab.ch
------
STL137@aol.com ===> Website: http://members.aol.com/stl137/
PGP keys: ~~~pgp.html Quotes: ~~~quotes.html
"I have sworn upon the altar of God eternal hostility against every form of
tyranny over the mind of man" - Thomas Jefferson
==============================================================================
From: Kurt Foster
Newsgroups: sci.math
Subject: Re: numbers, mirror numbers
Date: 20 Oct 1998 02:57:26 GMT
In <70d2bd$8mg$1@news01.btx.dtag.de>, Heidelberg said:
[snip]
. Take a number, Add the mirror number. Repeat this until you get a
. 'palindrom' - this is a number which looks the same from both sides,
. like 123431.
[snip]
. The problem: does this sequence stop for each number you start with ?
. Does it stop for 196 ? When ? -
.
Unknown, but the computations have been pushed pretty far. When I asked
about this in June 1996 I got the following response:
. From: cmcmanus
. Subject: Re: Status of palindrome conjecture?
. Date: 17 Jun 1996
. Message-ID: <4q4ged$q2p@news6.erols.com>
. references: <4q3s5k$9g7@natasha.rmii.com>
. newsgroups: sci.math
. Probably the final word on the palindromic quest involving 196 is found
. in two articles by John Walker.
. See:
www.fourmilab.ch/documents/threeyears/threeyears.html
. and
www.fourmilab.ch/documents/threeyears/two_months_more.html
. In the first paper, Walker reports on a 3-year project, which reached
. 2,415,836 steps, and a number containing 1 million digits, without ever
. yielding a palindrome.
. In the second paper, one Tim Irvin reports on a project building on
. Walker's efforts, and which took only two months to reach a nonpalindromic number of two
. million digits.
[snip]
==============================================================================
Mathematical Reviews on the Web 1999/01/13
Selected Matches for: Anywhere=(palindrom* and digit*)
_________________________________________________________________
Next Review
97m:11015 11A63
Yamashita, Koji
A problem on palindromic numbers. (English. English summary)
Math. Japon. 45 (1997), no. 1, 159--163.
For a positive integer $n$, written in base 10, let $n'$ be the
integer obtained by reversing the digits of $n$ and let $f\sb
1(n)=n+n'$. For $k\geq 2$, let $f\sb k(n)=f\sb 1(f\sb {k-1}(n))$. It
has been asked whether for each positive integer $n$ there exists a
positive integer $k$ for which $f\sb k(n)=(f\sb k(n))'$; that is, for
which $f\sb k(n)$ is palindromic. The author verifies that this is so
for $n\leq 195$. It is also shown that for all but perhaps 246 values
of $n<10\,000$, such a $k$ exists. For the exceptional cases, it is
shown that there is no such $k\leq 1000$.
Reviewed by H. L. Abbott
_________________________________________________________________
48 #5993 10A40
Trigg, Charles W.
Versum sequences in the binary system.
Pacific J. Math. 47 (1973), 263--275; corrections, ibid. 49 (1973),
619.
The result $S'$ of adding a positive integer $S$, relative to some
base, to the integer whose digits are those of $S$ reversed, is called
a versum. An old conjecture is that, relative to base 10, any
iterative sequence of versa $S,S',S",S"',\cdots$ contains a
palindromic number. D. C. Duncan [Sphinx 9 (1939), 91--92] showed that
the conjecture is false relative to base 2 by exhibiting a
palindromefree recursive cycle of 4 versums. The present author, also
working in base 2, is concerned with further instances of such
palindrome-free sequences as well as with palindromefree sequences not
exhibiting such recursive cycles.
Reviewed by J. B. Roberts
_________________________________________________________________
40 #7192 10.08
Gabai, Hyman; Coogan, Daniel
On palindromes and palindromic primes.
Math. Mag. 42 1969 252--254.
Let $N'$ be the integer obtained from the positive integer $N$ by
reversing the digits of $N$ written to the base $b$. Let $S\sb 1=N+N'$
and $S\sb {k+1}=S\sb k+S\sb k{}'$. An old conjecture states that, when
$b=10$, for each $N$ there is a $k=k(N)$ such that $S\sb k$ is
palindromic (i.e., $S\sb k=S\sb k{}'$). Whether there is such a $k$
for $N=196$ remains in doubt.
The authors point out that the conjecture fails when $b=2$. In fact,
when $N=837=1101000101$ the number $S\sb {4h}(N)$ is obtained from $N$
by replacing the third and eighth binary digits by $h+1$ zeros and
ones, respectively. This fact is enough to show that $S\sb k$ is
palindromic for no value of $k$. (This fact was communicated to the
reviewer in 1954 by T. S. Motzkin.)
When $b$ is ten, some interest has been expressed in the past in
palindromic primes. The authors have counted palindromic primes of 3,
5 and 7 digits and have found 15, 93 and 668. They note that
palindromic primes appear to be denser in the domain of all
palindromes than the ordinary primes are in the domain of all positive
integers.
Reviewed by D. H. Lehmer
© Copyright American Mathematical Society 1999
==============================================================================
[Comments:
1. There is a misprint in Brown's tables: 100(decimal) is a palindrome
base 3. The first presumably non-terminating number in base 3 is
103(decimal)=10211_{3}.
2. There appear to be plenty of other palindrome-free sequences.
Taking a few 5-digit integers at random, it is easy to find some
whose sequences appear not to terminate, yet which also appear not
to join the other sequences. For example, taken to 10,000 digits,
the given chains lead to numbers beginning
196:
1572252888008383803714884446997489576027567542194112240651342698432831
879:
1622485170768957736604537323008989748932494401998087270917988660255293
9988:
1459431769898319536260032364690389326185404937834733563462909955368107
while, for example, 32419 hits a 10,000-digit number at
1989576889978067474446699628914346714396644395657983240389351328928338
so it has not joined the sequence yet.
Likewise we try
36973, which hits 10000 digits at
1715082280172088855721474956398093581002933490396524032093400018750325
and 43844, which hits at
1590066149997562703886911665885975140825357757205142835710959608698026
So there appears to be little room for optimism that the sequences
coalesce.
Indeed, trying random integers shows that longer numbers are
increasingly likely to fail to reach a palindrome (testing to a few
tens of thousands of iterations).