Smarandache
On
Tl3Ilg Zhengping (Hanglhou Teacher's College,
Abstract
A
sequences and subsequences
Xu Kanghua Fuyang No.2 Middle Schoo~ Zhejiang, P. R. China)
Smarandache
sequence
partial
perfect
additive
sequence is studied completely in the first paragraph. In the second paragraph both Smarandache square-digital subsequence and squarepartial-digital subsequence are studied
Key
words
Smarandache subsequence.
ยง1 The be
a
Smarandache
square-digital
partial
perfect
subsequence, Smarandache
Smarandache partial
additive
square-partial-digital
perfect additive
Smarandache partial perfect additive
sequence
sequence,
is
sequence defined
to
sequence: 1, 1, 0, 2, -1, 1, 1, 3, - 2, 0, 0, 2, 0, 2, 2, 4,
-3,-1, -1, 1, -1, 1, 1, 3, -1, 1, ....
sequence has
This
P
j
i=1
and
p~1.
constructed in the fonowing way:
In [1]
function
for all
j=p+l
a 1p+1
(a)
that:
property
lp
La = La), It is
the
= ap +1 + 1
for all
M. Bencze raised
Can
you,
readers,
p
~ 1.
the following fmd
a
of n )?
146
two
general
questions: expression
of
an (as
Is it periodical or convergent or boWlded? (b) Please design (invent) yourselves other Smarandache (or partial perfect) f-sequences. In this paper we solved the question (a) completely. Suppose which
among
numbers
&l;
of
makes
8j
Thus
binary
the
&j
notation
=1,&, =0
of
as
n(n 2 2)
or 1 (i=O,l,路路路,k-l).
=l(i =0,1, .. 路,k),
g(n)
is
n
perfect
=(8A;8A;_1 "'8 8
Define
1
fen)
the
minimum
of
a(n) (i.e. a,,)
of
0
)2.'
are
the
i
that
as
the
=1. we
may
prove
the
expression
following:
r
1-
a(n) =
k, if k + 2f(n) + 2g(n) - 3 otherwise
We may
80
= 8 1 = '"
= 8):_1 = 0,
use mathematical induction to prove it
a(2) = 1,a(3) = 0= -1+ 2 x2+ 2 xO- 3= -1+ 2f(3) + 2g(3) -3.
So
the conclusion is valid for
Suppose
that
2,3"", n -len 2 3) .Let's
the
n =2,3,
conclusion
consider the
is
cases of
also
valid
for
n.
=k-l+l=k.
2
When
cases
should
not be
all
the
8 0,81> "',8):_1
discussed.
147
are
zeroes,
two
kinds
of
(1) If
Co
= o.
n
g(n) = g«EJc Elt:-l "'E1 EO )2) = g«E/cElt:-l "'E1 )2)+ 1 = g(i) + 1.
According
inductive hypothesis, we have n n n a(n) = a(-) + 1 =-(k-l)+ 2f(-) + 2g(-)- 3+ 1 to
2
2
2
=-k+ 2f(n) + 2(g(n) -1)-1 =-k+ 2f(n)+2g(n)-3. (2) If
= 1,
Co
three
subcases
exist
the
notation
[xl
(i) If E1 =0.
=f{~J+ 1), more
than
denotes
the
it's
integer
x.
g(n) = g«Et Et _1'''E1Eo)2) = g«&t&t-l '''E3&2 1)2) =
so,
greatest
easily
a(n) =
known from
g{~J+ 1) = o.
inductive hypothesis
aC[~J+ 1)-1 =-(k-1)+ 2f{~]+ 1)+ 2gC[~J+ 1) -3-1
=-k+ 2f(n)+ 2g(n)-3.
148
not
=(EkEk_1 "'Ei+ll~l' i
So,
tiIfIa
/{~J+l)=/(n)-i, g{~]+l)=i=i+g(n). According to
Then,
a(n) =
inductive hypothesis.
we
have
a<[~J+ 1)-1 =-(k-l)+ 2f{~J+ 1)+ 2g<[~J+ 1)-3-1
= -k+ 2(f(n)- i)+ 20 + g(n禄 - 3
=-k+ 2j(n) + 2g(n)-3.
j(n)=k+l, g(n)=O.
= (l~2'
SO
[~]+I=(e.te.t_l'路'E2el)2 +(1)2
from
1
k Ilma
Then
a{~]+I)-1 = k-l
a(n) =
= -k+ 2(k+ 1)+ 2 x 0- 3 = -k+ 2/(n) + 2g(n) -3. From the natural numbers
above,
conclusion is
the
(a)
for
all
the
n(n ~ 2).
Having proved above fact, question
true
can be solved
a(n) =k,
so sequence {a(n)}
periodical
and
easily. is
the
remaining problem in
For if
unbounded,
convergent.
149
n = 2,t,
we have
therefore
cannot be
§2
Smaranche square-digital subsequence and
Smaranche square-partial-digital subsequence The Smaranche subsequence:
square-digital
is
subsequence
defined
be
to
a
0, 1, 4, 9, 49, 100, 144, 400, 441,
terms and
all
are
digits
those
perfect
2 71 ,
we
choose
(Therefore
squares
only
the
0, 1,
only
4
9)
In [1] the
...,
0, 1, 4, 9, 16, 25, 36,
from
i.e.
form
MBencze
questioned:
where
NO···O
-.-'
Disregarding
is
N
also
a
the
square
perfect
square,
numbers how
of
many
2.t timu
other
numbers We
sequence
find by
In fact, the
belong
to
that
this
sequence?
1444, 11449,
calculating. we may
find
also
491401,
infinitely
many
numbers
that
to
the
belong
to
be
a
sequence.
(2·10 k +1)2
=
4·102.1: + 4·10 k + 1,
Smarandache sequence:
square-partial-digital
49, 100, 144, 169,
and
9 = 32, etc.).
In
the
same
nmnbers
of the
way
form
subsequence
~
1. is
other
it
be partitioned into groups squares (169 can be partitioned
18
questioned:
-.-' where
NO···O
nmnbers
to
can
N
is
also
belong
to
150
this
sequence?
a
of as
the
square
perfect
square,
Disregarding
2.t timG
many
de:fmed
400, 441, ...
i.e. the square nmnbers that digits which are also perfect 16 = 4 2
k
all
for
how
belong
We or
22
numbers
in
1he
-1
form
aOb (here
a
neither
zero).
is
b
find
may
10404, 11025, 11449, 11664, 40401, 41616, 42025, 43681, 93025, 93636, 161604, 164025, 166464, 251001, 254016, 259081, 363609, 491401, 641601, 646416, 813604, 819025. We
may
middle
the
701 2 ,
construct
infinitely
numbers
by
adcting
zero
in
of there numbers like 102 2 , 105 2 , 107 2 , 108 2 , 201\ 204 2 ,
8012 , 804 2 , 902 2 , 905 2
numbers
many
as
the
as
well.
we
may
find
some
other
following:
32436Q!, 102464Q1 25664Q1, !Q36324,
~Q64256,
36144144, 49196196,
81324324, 6425625§, 121484484, 169676676, 196784784, 484484121, 576576144, 676676169, 784784196 900900225, 1442401, 32436Q1 4004001
§3
Smaranche
cube-partial-digital
subsequence
1000, 8000, 10648, 27000, ... i.e. digits
the which
cube nmnbers that can be partitioned into groups are also perfect cubes (10648 can be partitioned
1=1\0=0\64=4\ and
8=2 3 )
Same question: disregarding M 0···0 IS also ............... ,where M
---
of as
the cube a perfect
numbers of the cube, how many
fonn: other
3k timG
numbers belong to this sequence? As the above said, we may find belong to the sequence as well
151
infmitely
many
numbers
that
for all for
example
21818127 =3033 ,
216648648216 =6006 3
,
21610801800! =6001 3 , !Q!8108216 =10063
â&#x20AC;˘
k~
0
REFERENCES 1. Bencze, M(I998). "Smarandache Relationships and Subsequences", Bulletin of Pure and Applied Sciences. Vol 17E (NO.1) 1998; P.55-62.
152