THE SMARANDACHE PERIODICAL SEQUENCES
by student M.R. Popov, Chandler College
1) Let N be a positive integer with not all digits the same,
and N' its digital reverse.
Then, let N1 = abs (N-N'), and N1' its digital reverse.
Again, let N2 = abs (N1-N1'), N2' its digital reverse,
and so on.
After a finite number of steps one finds an Nj which is equal
to a previous Ni, therefore the sequence is periodical
[because if N has, say, n digits, all other integers following
it will have n digits or less, hence their number is limited,
and one applies the Dirichlet's box principle].
For examples:
a) If one starts with N = 27, then N' = 72;
abs (27-72) = 45; its reverse is 54;
abs (45-54) = 09, ...
thus one gets: 27, 45, 09, 81, 63, 27, 45, ... ;
the Lenth of the Period LP = 5 numbers (27, 45, 09, 81, 63),
and the Lenth of the Sequence 'till the first repetition
occurs LS = 5 numbers either.
b) If one starts with 52, then one gets:
52, 27, 45, 09, 81, 63, 27, 45, ...;
then LP = 5 numbers, while LS = 6.
c) If one starts with 42, then one gets:
42, 18, 63, 27, 45, 09, 81, 63, 27, ...;
then LP = 5 numbers, while LS = 7.
For the sequences of integers of two digits, it seems like:
LP = 5 numbers (27, 45, 09, 81, 63; or a circular permutation
of them), and 5 <= LS <= 7.
Question 1: Find the Lenth of the Period (with its corresponding
numbers), and the Lenth of the Sequence 'till the first
repetition occurs for:
the integers of three digits,
and the integers of four digits.
(It's easier to write a computer program in these cases
to check the LP and LS.)
An example for three digits:
321, 198, 693, 297, 495, 099, 891, 693, ...;
(similar to the previous period, just inserting 9 in the
middle of each number).
Generalization for sequences of numbers of n digits.
2) Let N be a positive integer, and N' its digital reverse.
For a given positive integer c,
let N1 = abs (N'-c), and N1' its digital reverse.
Again, let N2 = abs (N1'-c), N2' its digital reverse,
and so on.
After a finite number of steps one finds an Nj which is equal
to a previous Ni, therefore the sequence is periodical
[same proof].
For eaxmple:
If N = 52, and c =1, then one gets:
52, 24, 41, 13, 30, 02, 19, 90, 08, 79, 96, 68, 85, 57, 74,
46, 63, 35, 52, ...;
thus LP = 18, LS = 18.
Question 2: Find the Lenth of the Period (with its corresponding
numbers), and the Lenth of the Sequence 'till the first
repetition occurs (with a given non-null c) for:
the integers of two digits,
and the integers of three digits.
(It's easier to write a computer program in these cases
to check the LP and LS.)
Generalization for sequences of numbers of n digits.
3) Let N be a positive integer with n digits a1a2...an, and c a
given integer > 1.
Multiply each digit ai of N by c, and replace ai with the last
digit of the product ai x c, say it is bi. Note N1 = b1b2...bn,
do the same procedure for N1, and so on.
After a finite number of steps one finds an Nj which is equal
to a previous Ni, therefore the sequence is periodical
[same proof].
For eaxmple:
If N = 68 and c = 7:
68, 26, 42, 84, 68, ...;
thus LP = 4, LS = 4.
Question 3: Find the Lenth of the Period (with its corresponding
numbers), and the Lenth of the Sequence 'till the first
repetition occurs (with a given c) for:
the integers of two digits,
and the integers of three digits.
(It's easier to write a computer program in these cases
to check the LP and LS.
Generalization for sequences of numbers of n digits.
4.1) Smarandache generalized periodical sequence:
Let N be a positive integer with n digits a1a2...an. If f is a
function defined on the set of integers with n digits or less,
and the values of f are also in the same set, then:
there exist two natural numbers i < j such that
f(f(...f(s)...)) = f(f(f(...f(s)...))),
where f occurs i times in the left side, and j times in the
right side of the previous equality.
Particularizing f, one obtains many periodical sequences.
Say:
If N has two digits a1a2, then: add 'em (if the sum is greater
than 10, add the resulted digits again), and substract 'em
(take the absolute value) -- they will be the first, and second
digit respectively of N1. And same procedure for N1.
Example: 75, 32, 51, 64, 12, 31, 42, 62, 84, 34, 71, 86, 52, 73,
14, 53, 82, 16, 75, ...
4.2) More General:
Let S be a finite set, and f : S ---> S a function. Then:
for any element s belonging to S, there exist two natural numbers
i < j such that
f(f(...f(s)...)) = f(f(f(...f(s)...))),
where f occurs i times in the left side, and j times in the right
side of the previous equality.
Reference:
F. Smarandache, "Sequences of Numbers", University of Craiova
Simposium of Students, December 1975.