Problems & Puzzles:
Conjectures
Conjecture
23. The
Shyam's conjecture about the Rare Numbers
Since 1989, Shyam Sunder Gupta from India has been
investigating the numbers RN (he names
them "rare numbers") such that:
* RN + RN' = A2
* RN - RN' = B2
* RN' is RN reversed (or vice versa)
* RN ¹
RN' (RN is not a palindrome)
Example: RN=65
because 65+56 = 11^2 & 65 - 56 = 3^2
The first 15 Rare Numbers RN
are:
65, 621770, 281089082, 2022652202,
2042832002,868591084757, 872546974178, 872568754178, 6979302951885,
20313693904202, 20313839704202, 20331657922202,
20331875722202,20333875702202, 40313893704200, ... (See
Neil's sequence A035519)
Shyam has demonstrated (*) that the RN's
have the following properties:
Let RN =abcd...mnpq
Then a, b, p & q can be only one of the
following alternatives, as a consequence of the digital properties of the
squares:
|
Digital
properties of RN |
| Property |
a |
q |
b & p |
| P1 |
2 |
2 |
b = p |
| P2 |
4 |
0 |
|b - p| = zero or even |
| P3 |
6 |
0 or 5 |
|b - p| = odd |
| P4 |
8 |
2 |
b + p = 9 |
| P5 |
8 |
3 |
b - p=7 or p - b = 3 |
| P6 |
8 |
7 |
b + p = 11 or b + p = 1 |
| P7 |
8 |
8 |
b = p |
From these properties the following corollaries
can be stated:
| Digital Corollaries
of RN |
|
C1
|
q can not be 1, 4,
6 & 9
|
|
C2
|
If q = 3 or 7, then the a = 8
|
|
C3
|
If a = q then (a = 2 or a = 8) and
b = p
|
|
C4
|
a-q = 0 or 1 or 4 or 5 or 6
|
|
C5
|
Digital root of RN = 2 or 5 or 8 or 9
|
Currently Shyam has found a total of 63 RN
numbers below 10^18, none of these ending in 3, five ending in 7
& none of these prime numbers.
Shyam wrote in his email dated the 4/2/2001:
It
is now conjectured that there does not exist any Rare number which is
also a Prime.
1)
Can this conjecture be proved or
disproved?
2) In case this is disproved, can anybody found a Rare Prime
number?
***
I would add the following additional open questions
about the RN numbers arising from the
Shyam's study and results:
OQ1: Are the quantity of RN
numbers infinite (**)?
OQ2: Find the first RN
ending in 3
OQ3: Can you develop an efficient algorithm
(***)
to find the RN
values?
(****)
Other properties of RN (sent by Shyam the
15/2/2001) are the following:
1) A must be divisible by 11, if RN consist of even number of
digits. So 121 must be a factor of A^2, if RN consist of even number of
digits.
2) B must be divisible by 11 if RN consist of odd number of
digits. Since B is always divisible by 3 , So B must be divisible by 33,
if RN consist of odd number of digits. Hence 1089 must be a factor of
B^2, if RN consist of odd number of digits.
It seems likely that odd RN's are fewer than even RN's. It is also
likely that RN's with odd number of digits are fewer than RN's with even
number of digits. So, odd RN's with odd number of digits are likely to
still fewer. In fact only one RN i.e. 6979302951885 is known up to 10^18
which is odd and also consist of odd number of digits... In view of
above , One additional Question: OQ4: Find the
second odd RN which consist of odd number of digits?.
______
Notes
(*) Systematic computations of rare numbers, Shyam Sunder Gupta,
The Mathematics Education, Vol. XXXII, No. 3, Sept. 1998.
(**) Palindrome Rare numbers are infinite. Shyam
has proved this showing at least one particular example, that is to say a
family of palindromes that are rare numbers: the first family is the
palindromes 2(0)k4(0)k2
(242, 20402, 2004002, etc.)
(***) Shyam wrote in his email: "I have developed a
computer program in Fortran to calculate Rare numbers. In fact with
refinement of the code over the years , the program has been made so
powerful that all numbers up to 10^14 can be just checked for Rare numbers
in less than a minute on Pentium III PC. In few hours I have been able to check up to 10^18."
(****) I would say that a radical improvement of
the efficacy of new algorithms must come from the discovery of new/more
digital properties of the RN's other than the published by Shyam.
So an important task for the interested reader is to go deeper in the
mathematical properties of these numbers.
Solution
Well, this conjecture is harder than I thought.
As a matter of fact no one contribution has been done since posted. But
Shyam continues working on the issue and got two more solutions, sent the
1/6/01:
"I am pleased to convey
some new results which answers Q.4 of the Conjecture 23. The following two
Rare Numbers (RN) of odd number of digits (i.e. 19 digits) are odd.
6531727101458000045
8200756128308135597
Both the above Rare Numbers are non-prime so the conjecture still
holds good."
***
Shyam wrote on December 11, 2019:
OQ2: Find the first RN ending in 3
I am pleased to report that first Rare Number (RN)ending
in 3 was found by me few years back but I missed
it to report earlier due to time constraints. The First
RN ending in 3 consist of 22 digit.
RN = 8888070771864228883913 RN' = 3193888224681770708888
RN + RN' = 12081958996545999592801 = 109917964849^2
RN - RN' = 5694182547182458175025 = 75459807495^2
I also report that There are 124 Rare numbers below
10^22. All Rare numbers found below 10^22 are non-primes so the
conjecture still holds good.
***
Metin Sariyar wrote on Dec 19, 2019:
Q1:
I noticed that, All
numbers m = a^2 + b^2 such that reversal(m) = 2*a*b is a term . For
the numbers with this property, m-reversal(m) = (a-b)^2 and m +
reversal(m) = (a+b)^2
Some terms with this property: 65, 621770, 2042832002, 872546974178,
872568754178, 20313693904202,... If it is proved that the numbers
with this property are infinitely many than the sequence is
infinite.
but for the sequence
R + R' = A3
R - R' = B3
R'=Reversal(R)
The
sequence is infinite. Proof : The numbers of the form
R=3*10^2n+1 for n>=1 are terms (1030, 1000300,
1000003000, ... ) (but these may not be the all the
terms of this seq.) then R'=10^3n+3*10^n for n>=1 . So,
R-R'=10^3n+3*10^n-3*10^2n-1=(10^n-1)^3
(9^3,99^3,999^3,9999^3,99999^3,...) and R+R'=10^3n+3*10^n+3*10^2n+1=(10^n+1)^3
(11^3, 101^3, 1001^3,...)
***
|
