SOME NOTIONS AND QUESTIONS IN NUMBER THEORY
Vol. II
edited by
Mihaly Bencze Lucian Tutescu
6, Harmanului Street Mathematics Department
2212 Sacele 3 Fratii Buzesti College
Jud. Brasov, Romania Craiova, R-1100, Romania
The second volume of Smarandache sequences, following that edited
by the late C. Dumitrescu and V. Seleacu from the University of Craiova,
is been published by the American Research Press.
We want to thank Dr. M. L. Perez and J. Castillo for their help and
encouragements in this action.
1) Smarandache Concatenate Natural Sequence:
1,22,333,4444,55555,666666,7777777,88888888,999999999,
10101010101010101010,1111111111111111111111,
121212121212121212121212,13131313131313131313131313,
1414141414141414141414141414,151515151515151515151515151515,...
2) Smarandache Concatenated Prime Sequence:
2, 23, 235, 2357, 235711, 23571113, 2357111317, 235711131719,
23571113171923, ...
Smarandache Back Concatenated Prime Sequence:
2, 32, 532, 7532, 117532, 13117532, 1713117532, 191713117532,
23191713117532, ...
Conjecture: there are infinitely many primes among the first
sequence numbers!
3) Smarandache Concatenated Odd Sequence:
1, 13, 135, 1357, 13579, 1357911, 135791113, 13579111315,
1357911131517, ...
Smarandache Back Concatenated Odd Sequence:
1, 31, 531, 7531, 97531, 1197531, 131197531, 15131197531,
1715131197531, ...
Conjecture: there are infinitely many primes among these numbers!
4) Smarandache Concatenated Even Sequence:
2, 24, 246, 2468, 246810, 24681012, 2468101214, 246810121416, ...
Smarandache Back Concatenated Even sequence:
2, 42, 642, 8642, 108642, 12108642, 1412108642, 161412108642, ...
Conjecture: none of them is a perfect power!
5) Smarandache Concatenated S-Sequence {generalization}:
Let s1, s2, s3, s4, ..., sn, ... be an infinite integer sequence
(noted by S).
Then:
____ ______ ________ _____________
s1, s1s2, s1s2s3, s1s2s3s4, s1s2s3s4...sn, ...
is called the Concatenated S-sequence,
____ ______ ________ _____________
s1, s2s2, s3s2s1, s4s3s2s1, sn...s4s3s2s1, ...
is called the Back Concatenated S-sequence.
Questions: a) How many terms of the Concatenated S-sequence belong
to the initial S-sequence?
b) Or, how many terms of the Concatenated S-sequence
verify the relation of other given sequences?
The first three cases are particular.
Look now at some other examples, when S is the sequence of squares,
cubes, Fibonacci respectively (and one can go so on):
Smarandache Concatenated Square Sequence:
1, 14, 149, 14916, 1491625, 149162536, 14916253649, 1491625364964, ...
Smarandache Back Concatenated Square Sequence:
1, 41, 941, 16941, 2516941, 362516941, 49362516941, 6449362516941, ...
How many of them are perfect squares?
Smarandache Concatenated Cubic Sequence:
1, 18, 1827, 182764, 182764125, 182764125216, 182764125216343, ...
Smarandache Back Concatenated Cubic Sequence:
1, 81, 2781, 642781, 125642781, 216125642781, 343216125642781, ...
How many of them are perfect cubes?
Smarandache Concatenated Fibonacci Sequence:
1, 11, 112, 1123, 11235, 112358, 11235813, 1123581321, 112358132134, ...
Smarandache Back Concatenated Fibonacci Sequence:
1, 11, 211, 3211, 53211, 853211, 13853211, 2113853211, 342113853211, ...
Does any of these numbers is a Fibonacci number?
6) Smarandache Power Function:
SP(n) is the smallest number m such that m^m is
divisible by n.
The following sequence SP(n) is generated:
1, 2, 3, 2, 5, 6, 7, 4, 3, 10, 11, 6, 13, 14, 15, 4, 17, 6, 19, 10,
21, 22, 23, 6, 5, 26, 3, 14, 29, 30, 31, 4, 33, 34, 35, 6, 37,38,
39, 20, 41, 42, ...
Remarks:
If p is prime, then SP(p) = p.
If r is square free, then SP(r) = r.
If n = (p ^ s )x...x(p ^ s ) and all s <= p , then SP(n) =n.
1 1 k k i i
If n = p^s, where p is prime, then:
p, if 1 <= s <= p;
SP(n) =
p^2, if p+1 <= s <= 2p^2;
p^3, if 2p^2+1 <= s <= 3p^3;
..................................
p^t, if (t-1)p^(t-1) <= s <= tp^t .
Reference:
F. Smarandache, "Collected Papers", Vol. III, Tempus Publ.Hse.,
Bucharest, 1998.
7) Smarandache Reverse Sequence:
1,21,321,4321,54321,654321,7654321,87654321,987654321,10987654321,
1110987654321,121110987654321,...
8) Smarandache Multiplicative Sequence:
2,3,6,12,18,24,36,48,54,...ิ
General definition: if m , m , are the first two terms ofthe sequence,
1 2
then m , for k >= 3, is the smallest number equal to the product of two
k
previous distinct terms.
All terms of rank >= 3 are divisible by m , and m .
1 2
In our case the first two terms are 2, respectively 3.
9) Smarandache Wrong Numbers:
_________
(A number n = a a ...a , of at least two digits, with theุ
1 2 k
property:
the sequence a , a , ..., a , b , b , ... (where b is the product
1 2 k k+1 k+2 k+i
of the previous k terms, for any i >= 1) contains n as itsterm.)
The author conjectured that there is no Smarandache wrongnumber (!)
Therefore, this sequence is empty.
10) Smarandache Impotent Numbers:
2,3,4,5,7,9,11,13,17,19,23,25,29,31,37,41,43,47,49,53,59,61,...
(A number n those proper divisors product is less than n.)
2
Remark: this sequence is { p, p ; where p is a positiveprime }.
11) Smarandache Random Sieve:
1,5,6,7,11,13,17,19,23,25,29,31,35,37,41,43,47,53,59,...
General definition:
- choose a positive number u at random;
1
- delete all multiples of all its divisors, except this number;
- choose another number u greater than u among those remaining;
2 1
- delete all multiples of all its divisors, except this second number;
... and so on.
The remaining numbers are all coprime two by two.
The sequence obtained u , k >= 1, is less dense than theprime number
k
sequence, but it tends to the prime number sequence as ktends to infinite.
That's why this sequence may be important.
In our case, u = 6, u = 19, u = 35, ... .
1 2 3
12) Smarandache Square Base:
0,1,2,3,10,11,12,13,20,100,101,102,103,110,111,112,1000,1001,1002,1003,
1010,1011,1012,1013,1020,10000,10001,10002,10003,10010,10011,10012,10013,
10020,10100,...
(Each number n written in the Smarandache square base.)
(Smarandache defined over the set of natural numbers the following infinite
base: for k >= 1 s = k^2.)
k
He proved that every positive integer A may be uniquely written in
the Smarandache square base as:
n
___________ def ---
A = (a ... a a ) === \ a s , with
n 2 1 (C2) / i i
---
i=1
0 <= a <= 3, 0 <= a <= 2, and 0 <= a <= 1, for i >= 3,
1 2 i
and of course a = 1, in the following way:
n
- if s <= A < s then A = s + r ;
n n+1 n 1
- if s <= r < s then r = s + r , m < n;
m 1 m+1 1 m 2
and so on untill one obtains a rest r = 0.
j
Therefore, any number may be written as a sum of squares (1 not counted
as a square -- being obvious) + e, where e = 0, 1, 2, or 3.
If we note by s(A) the Smarandache Superior Square Part of A (i.e. the
largest square less than or equal to A), then A is written in the
Smarandache square base as:
A = s(A) + s(A-s(A)) + s(A-s(A)-s(A-s(A))) + ... .
This base may be important for partitions with squares.
13) Smarandache Cubic Base:
0,1,2,3,4,5,6,7,10,11,12,13,14,15,16,17,20,21,22,23,24,25,26,27,30,31,32,
100,101,102,103,104,105,106,107,110,111,112,113,114,115,116,117,120,121,
122,123,...
(Each number n written in the Smarandache cubic base.)
(Smarandache defined over the set of natural numbers the following infinite
base: for k >= 1 s = k^3.)
k
He proved that every positive integer A may be uniquelywritten in
the Smarandache cubic base as:
n
___________ def ---
A = (a ... a a ) === \ a c , with
n 2 1 (C3) / i i
---
i=1
0 <= a <= 7, 0 <= a <= 3, 0 <= a <= 2, and 0 <= a <= 1 for i >= 4,
1 2 3 i
and of course a = 1, in the following way:
n
- if c <= A < c then A = c + r ;
n n+1 n 1
- if c <= r < c then r = c + r , m < n;
m 1 m+1 1 m 2
and so on untill one obtains a rest r = 0.
j
Therefore, any number may be written as a sum of cubes (1 notcounted
as a cube -- being obvious) + e, where e = 0, 1, ..., or 7.
If we note by c(A) the Smarandache superior square part of A(i.e. the
largest cube less than or equal to A), then A is written inthe
Smarandache cube base as:
A = c(A) + c(A-c(A)) + c(A-c(A)-c(A-c(A))) + ... .
This base may be important for partitions with cubes.
14) Smarandache Triangular Base:
1,2,10,11,12,100,101,102,110,1000,1001,1002,1010,1011,10000,
10001,10002,10010,10011,10012,100000,100001,100002,100010,
100011,100012,100100,1000000,1000001,1000002,1000010,1000011,
1000012,1000100,...
( Numbers written in the triangular base, defined
as follows: t(n) = n(n+1)/2, for n >= 1. )
15) Smarandache Double Factorial Base:
1,10,100,101,110,200,201,1000,1001,1010,1100,1101,1110,1200,
10000,10001,10010,10100,10101,10110,10200,10201,11000,11001,
11010,11100,11101,11110,11200,11201,12000,...
( Numbers written in the double factorial base,
defined as follows: df(n) = n!! )
16) Smarandache General Base:
Let 1 = b1 < b2 < b3 < ... < bn < ... be an integer strictly
increasing sequence. Then any positive integer can uniquely be
written in this infinite base
(in the same way as the previous particular bases).
We consider the
If we note by b(A) the Smarandache Superior Square Part of A (i.e. the
largest term bk less than or equal to A), then A is written in the
Smarandache square base as:
A = b(A) + b(A-b(A)) + b(A-b(A)-b(A-b(A))) + ... .
This general base may be important for partitions with increasing
sequences.
Reference:
F. Smarandache, "Properties of the Numbers", University ofCraiova
Archives, 1975;
[see also Arizona State University Special Collections,Tempe,
Arizona, USA]
17) Smarandache Non-Multiplicative Sequence:
General definition: let m , m , ..., m be the first k giventerms
1 2 k
of the sequence, where k >= 2;
then m , for i >= k+1, is the smallest number not equal to the product of
i
k previous distinct terms.
18) Smarandache Non-Arithmetic Progression:
1,2,4,5,10,11,13,14,28,29,31,32,37,38,40,41,64,...
General definition: if m , m , are the first two terms ofthe sequence,
1 2
then m , for k >= 3, is the smallest number such that no3-term
k
arithmetic progression is in the sequence.ิ
In our case the first two terms are 1, respectively 2.
Generalization: same initial conditions, but no i-termarithmetic
progression in the sequence ( for a given i >= 3 ).
19) Smarandache Prime Product Sequence:
2,7,31,211,2311,30031,510511,9699691,223092871,6469693231,200560490131,
7420738134811,304250263527211,...
P = 1 + p p ...p , where p is the k-th prime.
n 1 2 n k
Question: How many of them are prime?
20) Smarandache Square Product Sequence:
2,5,37,577,14401,518401,25401601,1625702401,131681894401,13168189440001,
1593350922240001,...
S = 1 + s s ...s , where s is the k-th square number.
n 1 2 n k
Question: How many of them are prime?
21) Smarandache Cubic Product Sequence:
2,9,217,13825,1728001,373248001,128024064001,65548320768001,...
C = 1 + c c ...c , where c is the k-th cubic number.
n 1 2 n k
Question: How many of them are prime?
22) Smarandache Factorial Product Sequence:
2,3,13,289,34561,24883201,125411328001,5056584744960001,...
F = 1 + f f ...f , where f is the k-th factorial number.
n 1 2 n k
Question: How many of them are prime?
23) Smarandache U-Product Sequence {generalization}:
Let u , n >= 1, be a positive integer sequence. Then we define
n
a sequence as follows:
U = 1 + u u ...u .
n 1 2 n
Reference:
F. Smarandache, "Properties of the Numbers", University ofCraiova
Archives, 1975;
[see also Arizona State University Special Collections,Tempe,
Arizona, USA].
24) Smarandache Sequences of Sub-Sequences:
For each of the following 7 sequences:
a) Smarandache Crescendo Sub-Sequences:
1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, 1, 2, 3,4, 5, 6,
1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 8, . . .
b) Smarandache Decrescendo Sub-sequences:
1, 2, 1, 3, 2, 1, 4, 3, 2, 1, 5, 4, 3, 2, 1, 6, 5, 4,3, 2, 1,
7, 6, 5, 4, 3, 2, 1, 8, 7, 6, 5, 4, 3, 2, 1, . . .
c) Smarandache Crescendo Pyramidal Sub-sequences:
1, 1, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 4, 3, 2, 1,
1, 2, 3, 4, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, . . .
d) Smarandache Decrescendo Pyramidal Sub-sequences:
ิ
5, 4, 3, 2, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 2, 3, 4, 5, 6, . . .
e) Smarandache Crescendo Symmetric Sub-sequences:
1, 1, 1, 2, 2, 1, 1, 2, 3, 3, 2, 1, 1, 2, 3, 4, 4, 3, 2,1,
1, 2, 3, 4, 5, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 6, 5, 4, 3, 2,1, . . .
f) Smarandache Decrescendo Symmetric Sub-sequences:
1, 1, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 4, 3, 2, 1, 1, 2, 3,4,
5, 4, 3, 2, 1, 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1, 1, 2, 3, 4, 5,6, . . .
g) Smarandache Permutation Sub-sequences:
1, 2, 1, 3, 4, 2, 1, 3, 5, 6, 4, 2, 1, 3, 5, 7, 8, 6, 4,2,
1, 3, 5, 7, 9, 10, 8, 6, 4, 2, 1, 3, 5, 7, 9, 10, 8, 6, 4, 2,. . .
Find a formula for the general term of each sequence.
Solutions:
For purposes of notation in all problems, let
a(n)
denote the n-th term in the complete sequence and
b(n)
the n-th subsequence. Therefore, a(n) will be a number and b(n) a
sub-sequence.
a) Clearly, b(n) contains n terms. Using a well-known summation formula,
at the end of b(n) there would be a total of
n(n+1)/2
terms.
Therefore, since the last number of b(n) is n,
a((n(n+1))/2) = n.
Finally, since this would be the terminal number in the sub-sequence
b(n) = 1, 2, 3, . . . , n
the general formula is
a(((n(n+1))/2) - i) = n - i
for n >= 1 and 0 <= i <= n - i.
b) With modifications for decreasing rather than increasing, theproof
is essentially the same. The final formula is
a(((n(n+1))/2) - i) = 1 + i
for n >= 1 and 0 <= i <= n - 1.
7
c) Clearly, b(n) has 2n - 1 terms. Using the well-known formulaof
summation
1 + 3 + 5 + . . . + (2n - 1) = n^2.
the last term of b(n) is in position n^2 and a(n^2) = 1. Thelargest
number in b(n) is n, so counting back n - 1 positions, theyincrease in value
by one each step until n is reached.
a(n^2 - i) = 1 + i, for 0 <= i <= n-1.
After the maximum value at n-1 positions back from n^2, thevalues decrease
by one. So at the nth position back, the value is n-1, at the(n-1)st
position back the value is n-2 and so forth.
a(n^2 - n - i) = n - i - 1
for 0 <= i <= n - 2.
d) Using similar reasoning
2
a(n ) = n for n >= 1
and
a(n^2 - i) = n - i, for 0 <= i <= n-1
a(n^2 - n - i) = 2 + i, for 0 <= i <= n-2.
e) Clearly, b(n) contains 2n terms. Applying another well-known summation
formula
2 + 4 + 6 + . . . + 2n = n(n+1), for n >= 1.
Therefore, a(n(n+1)) = 1. Counting backwards n-1 positions, eachterm
decreases by 1 up to a maximum of n.
a((n(n+1))-i) = 1 + i, for 0 <= i <= n-1.
The value n positions down is also n and then the terms decreaseby one
back down to one.
a((n(n+1))-n-i) = n - i, for 0 <= i <= n-1.
f) The number of terms in b(n) is the same as that for (e). Theonly
difference is that now the direction of increase/decrease isreversed.
a((n(n+1))-i) = n - i, for 0 <= i <= n-1.
a((n(n+1))-n-i) = 1 + i, for 0 <= i <= n-1.
g) Given the following circular permutation on the first
n integers.
| 1 2 3 4 . . . n-2 n-1 n |
phi = | |
n | 1 3 5 7 . . . 6 4 2 |
Once again, b(n) has 2n terms. Therefore,
a(n(n+1)) = 2.
Counting backwards n-1 positions, each term is two larger thanthe successor
a((n(n+1))-i) = 2 + 2i, for 0 <= i <= n-1.
The next position down is one less than the previous and afterthat, each
term is again two less the successor.
a((n(n+1))-n-i) = 2n - 1 - 2i, for 0 <= i <= n-1.
As a single formula using the permutation
a((n(n+1)-i) = phi (2n-i), for 0 <= i <= 2n-1.
n
F. Smarandache, "Collected Papers", Vol. II, Tempus Publ. Hse.,Bucharest,
1996.
F. Smarandache, "Numerical Sequences", University of Craiova,1975;
[ See Arizona State University, Special Collection, Tempe, AZ,USA ].
2
25) Smarandache function (numbers):
1, 2, 3, 2, 5, 6, 7, 4. 3, 10, 11, 6, 13, 14, 15, 4, 17, 6,19, 10,
21, 22, 23, 12, 5, 26, 9, 14, 29, 30, 31, 8, 33, ...
2
( S2(n) is the smallest integer m such that m is divisible by n)
3
26) Smarandache function (numbers):
1, 2, 3, 2, 5, 6, 7, 8, 3, 10, 11, 6, 13, 14, 15, 4, 17, 6,19, 10,
21, 22, 23, 6, 5, 26, 3, 14, 29, 30, 31, 4, 33, ...
3
( S3(n) is the smallest integer m such that m is divisible by n)
27) Smarandache Anti-Symmetric Sequence:
11,1212,123123,12341234,1234512345,123456123456,12345671234567,
1234567812345678,123456789123456789,1234567891012345678910,
12345678910111234567891011,123456789101112123456789101112,...
28-36) Smarandache-Recurrence Type Sequences:
A. 1,2,5,26,29,677,680,701,842,845,866,1517,458330,458333,
458354,...
( ss2(n) is the smallest number, strictly greater than the
previous one, which is the squares sum of two previous
distinct terms of the sequence;
in our particular case the first two terms are 1 and 2. )
Recurrence definition:
1) The number a belongs to SS2;
2) If b, c belong to SS2, then b^2 + c^2 belongs to SS2too;
3) Only numbers, obtained by rules 1) and/or 2) applied a
finite number of times, belong to SS2.
The sequence (set) SS2 is increasingly ordered.
[ Rule 1) may be changed by: the given numbers a1, a2,...,
ak , where k >= 2, belong to SS2. ]
B. 1,1,2,4,5,6,16,17,18,20,21,22,25,26,27,29,30,31,36,37,38,40,41,
42,43,45,46,...
( ss1(n) is the smallest number, strictly greater than
the previous one (for n >= 3), which is the squares sum ofone
or more previous distinct terms of the sequence;
in our particular case the first term is 1. )
Recurrence definition:
1) The number a belongs to SS1;
2) If b1, b2, ..., bk belong to SS1, where k >= 1, then
b1^2 + b2^2 + ... + bk^2 belongs to SS1 too;
3) Only numbers, obtained by rules 1) and/or 2) applied a
finite number of times, belong to SS1.
The sequence (set) SS1 is increasingly ordered.
[ Rule 1) may be changed by: the given numbers a1, a2,...,
ak , where k >= 1, belong to SS1. ]
C. 1,2,3,4,6,7,8,9,11,12,14,15,16,18,19,21,...
( nss2(n) is the smallest number, strictly greater than the
previous one, which is NOT the squares sum of two previous
distinct terms of the sequence;
in our particular case the first two terms are 1 and 2. )
Recurrence definition:
1) The numbers a <= b belong to NSS2;
2) If b, c belong to NSS2, then b^2 + c^2 DOES NOT belongto
NSS2; any other numbers belong to NSS2;
3) Only numbers, obtained by rules 1) and/or 2) applied a
finite number of times, belong to NSS2.
The sequence (set) NSS2 is increasingly ordered.
[ Rule 1) may be changed by: the given numbers a1, a2,...,
ak , where k >= 2, belong to NSS2. ]
D. 1,2,3,6,7,8,11,12,15,16,17,18,19,20,21,22,23,24,25,26,27,28,
29,30,31,32,33,34,35,38,39,42,43,44,47,...
( nss1(n) is the smallest number, strictly greater than the ิ
previous distinct terms of the sequence;
in our particular case the first term is 1. )
Recurrence definition:
1) The number a belongs to NSS1;
2) If b1, b2, ..., bk belong to NSS1, where k >= 1, then
b1^2 + b2^2 + ... + bk^2 DO NOT belong to NSS1;
any other numbers belong to NSS1;
3) Only numbers, obtained by rules 1) and/or 2) applied afinite
number of times, belong to NSS1.
The sequence (set) NSS1 is increasingly ordered.
[ Rule 1) may be changed by: the given numbers a1, a2,...,
ak , where k >= 1, belong to NSS1. ]
E. 1,2, 9, 730,737, 389017001,389017008,389017729,...
( cs2(n) is the smallest number, strictly greater than the
previous one, which is the cubes sum of two previous
distinct terms of the sequence;
in our particular case the first two terms are 1 and 2. )
Recurrence definition:
1) The numbers a <= b belong to CS2;
2) If c, d belong to CS2, then c^3 + d^3 belongs to CS2too;
3) Only numbers, obtained by rules 1) and/or 2) applied a
finite number of times, belong to CS2.
The sequence (set) CS2 is increasingly ordered.
[ Rule 1) may be changed by: the given numbers a1, a2,...,
ak , where k >= 2, belong to CS2. ]
F. 1,1, 2, 8,9,10, 512,513,514,520,521,522,729,730,731,737,738,
739,1241,...
( cs1(n) is the smallest number, strictly greater than
the previous one (for n >= 3), which is the cubes sum of one
or more previous distinct terms of the sequence;
in our particular case the first term is 1. )
Recurrence definition:
1) The number a belongs to CS1.
2) If b1, b2, ..., bk belong to CS1, where k >= 1, then
b1^3 + b2^3 + ... + bk^3 belongs to CS2 too.
3) Only numbers, obtained by rules 1) and/or 2) applied a
finite number of times, belong to CS1.
The sequence (set) CS1 is increasingly ordered.
[ Rule 1) may be changed by: the given numbers a1, a2,..., ิ
G. 1,2,3,4,5,6,7,8,10,11,12,13,14,15,16,17,18,19,20,21,22,23,
24,25,26,27,29,30,31,32,33,34,36,37,38,...
( ncs2(n) is the smallest number, strictly greater than the
previous one, which is NOT the cubes sum of two previous
distinct terms of the sequence;
in our particular case the first two terms are 1 and 2. )
Recurrence definition:
1) The numbers a <= b belong to NCS2;
2) If c, d belong to NCS2, then c^3 + d^3 DOES NOT belongto
NCS2; any other numbers do belong to NCS2;
3) Only numbers, obtained by rules 1) and/or 2) applied a
finite number of times, belong to NCS2.
The sequence (set) NCS2 is increasingly ordered.
[ Rule 1) may be changed by: the given numbers a1, a2,...,
ak , where k >= 2, belong to NCS2. ]
H. 1,2,3,4,5,6,7,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,
25,26,29,30,31,32,33,34,37,38,39,...
( ncs1(n) is the smallest number, strictly greater than
the previous one, which is NOT the cubes sum
of one or more previous distinct terms of the sequence;
in our particular case the first term is 1. )
Recurrence definition:
1) The number a belongs to NCS1.
2) If b1, b2, ..., bk belong to NCS1, where k >= 1, then
b1^2 + b2^2 + ... + bk^2 DO NOT belong to NCS1.
3) Only numbers, obtained by rules 1) and/or 2) applied afinite
number of times, belong to NCS1.
The sequence (set) NCS1 is increasingly ordered.
[ Rule 1) may be changed by: the given numbers a1, a2,...,
ak , where k >= 1, belong to NCS1. ]
I. Smarandache-General-Recurrence (Positive) Type Sequence:
General (positive) recurrence definition:
Let k >= j be natural numbers, and a1, a2, ..., ak given
elements, and R a j-relationship (relation among j elements).
Then:
1) The elements a1, a2, ..., ak belong to SGPR.
2) If m1, m2, ..., mj belong to SGPR, then
R ( m1, m2, ..., mj ) belongs to SGPR too.
3) Only elements, obtained by rules 1) and/or 2) applied a finite
number of times, belong to SGPR.
The sequence (set) SGPR is increasingly ordered.
Method of construction of the Smarandache (positive) general recurrence
sequence:
- level 1: the given elements a1, a2, ..., ak belong to SGPR;
- level 2: apply the relationship R for all combinations of
j elements among a1, a2, ..., ak;
the results belong to SGPR too;
order all elements of levels 1 and 2 together;
............................................................
- level i+1:
if b1, b2, ..., bm are all elements of levels 1, 2,..., i-1,
and c1, c2, ..., cn are all elements of level i,
then apply the relationship R for all combinations of j
elements among b1, b2, ..., bm, c1, c2, ..., cn such that at
least a element is from the level i;
the results belong to SRPG too;
order all elements of levels i and i+1 together;
and so on ...
J. Smarandache-General-Recurrence (Negative) Type Sequence:
General (negative) recurrence definition:
Let k >= j be natural numbers, and a1, a2, ..., ak given
elements, and R a j-relationship (relation among j elements).
Then:
1) The elements a1, a2, ..., ak belong to SGNR.
2) If m1, m2, ..., mj belong to SGNR, then
R ( m1, m2, ..., mj ) does NOT belong to SGNR;
any other elements do belong to SGNR.
3) Only elements, obtained by rules 1) and/or 2) applied a finite
number of times, belong to SGNR.
The sequence (set) SGNR is increasingly ordered.
Method of construction of the Smarandache (negative) general recurrence
sequence:
- level 1: the given elements a1, a2, ..., ak belong to SGNR;
- level 2: apply the relationship R for all combinations of
j elements among a1, a2, ..., ak;
NONE of the results belong to SGNR;
the smallest element, strictly greater than all a1, a2, ..., ak,
and different from the previous results, belongs to SGNR;
order all elements of levels 1 and 2 together;
............................................................
- level i+1:
if b1, b2, ..., bm are all elements of levels 1, 2,..., i-1,
and c1, c2, ..., cn are all elements of level i,
then apply the relationship R for all combinations of j
elements among b1, b2, ..., bm, c1, c2, ..., cn such that at
least a element is from the level i;
NONE of the results belong to SRNG;
the smallest element, strictly greater than all previous elements,
and different from the previous results, belongs to SGNR;
order all elements of levels i and i+1 together;
and so on ...
37-39) Smarandache-Partition Type Sequences:
A. 1,1,1,2,2,2,2,3,4,4,...
( How many times is n written as a sum of non-null squares,
disregarding the terms order;
for example:
9 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2
= 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2
= 1^2 + 2^2 + 2^2
= 3^2,
therefore ns(9) = 4. )
B. 1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,4,4,4,5,5,
5,5,5,6,6,...
( How many times is n written as a sum of non-null cubes,
disregarding the terms order;
for example:
9 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3
= 1^3 + 2^3,
therefore nc(9) = 2. )
C. Smarandache-General-Partition Type Sequence:
{ How many times can n be written under the form:
n = R ( f(n1), f(n2), ..., f(nk) )
for some k and n1, n2, ..., nk between 1 and n ? }
[ Particular cases when f(x) = x^2, or x^3, or x!, or x^x, etc.
and the relation R is the abvious addition of numbers, or
multiplication, etc. ]
Reference:
F. Smarandache, "Properties of the Numbers", University ofCraiova
Archives, 1975; ( see also Arizona State University, Special
Collection, Tempe, AZ, USA ).
40) Smarandache Reverse Sequence:
1,21,321,4321,54321,654321,7654321,87654321,987654321,10987654321,
1110987654321,121110987654321,13121110987654321,1413121110987654321,
151413121110987654321,16151413121110987654321,...
41) Smarandache Multiplicative Sequence:
2,3,6,12,18,24,36,48,54,72,96,108,...
General definition: if m , m , are the first two terms ofthe sequence,
1 2
then m , for k >= 3, is the smallest number equal to theproduct of two
k
previous distinct terms.
All terms of rank >= 3 are divisible by m , and m .
1 2
a b t
All terms of rank >= 3 have the canonical form: d d ... d ,
1 2 r
where d , d , ..., d are the prime distinct divisors of m m.
1 2 r 12ิ
42) Smarandache Wrong Numbers:
_________
(A number n = a a ...a , of at least two digits, with thefollowing
1 2 k
property:
the sequence a , a , ..., a , b , b , ... (where b is the product
1 2 k k+1 k+2 k+i
of the previous k terms, for any i >= 1) contains n as itsterm.)
The author conjectured that there is no Smarandache wrongnumber (!)
Therefore, this sequence is empty.
43) Smarandache Impotent Numbers:
2,3,4,5,7,9,11,13,17,19,23,25,29,31,37,41,43,47,49,53,59,61,67,71,73,
79,83,89,97,101,103,107,109,113,121,127,131,137,139,149,151,157,163,
167,169,173,179,181,191,193,197,199,211,223,...
(A number n those proper divisors product is less than n.)
2
Remark: This sequence is the set { p, p ; where p is apositive
prime }.
44) Smarandache Non-Geometric Progression:
1,2,3,5,6,7,8,10,11,13,14,15,16,17,19,21,22,23,24,26,27,29,30,31,33,34,35,
37,38,39,40,41,42,43,45,46,47,48,50,51,53,...
General definition: if m , m , are the first two terms ofthe sequence,
1 2
then m , for k >= 3, is the smallest number such that no 3-term
k
geometric progression is in the sequence.
In our case the first two terms are 1, respectively 2.
Generalization: if m , m , ..., m are the first i-1 terms of the
1 2 i-1
sequence, i >= 3, then m , for k >= i, is the smallest number such
k
that no i-term geometric progression is in the sequence.
45) Smarandache Unary Sequence:
11,111,11111,1111111,11111111111,1111111111111,11111111111111111,
1111111111111111111,11111111111111111111111,111111111111111111111ิ
1111111111111111111111111111111,...
______
u(n) = 11...1 , p digits of "1", where p is the n-thprime.
n n
The old question: are there an infinite number of primesbelonging to
the sequence?
46) Smarandache No-prime-digits Sequence:
1,4,6,8,9,10,11,1,1,14,1,16,1,18,19,0,1,4,6,8,9,0,1,4,6,8,9,40,41,42,4,44,
4,46,4,48,49,0,...
(Take out all prime digits of n.)
Is it any number that occurs infinetely many times in thissequence?
(for example 1, or 4, or 6, or 11, etc.).
Solution by Dr. Igor Shparlinski,
It seems that: if, say n has already occured, then for example
n3, n33, n333, etc. gives infinitely many repetitions of this
number.
47) Smarandache No-square-digits Sequence:
2,3,5,6,7,8,2,3,5,6,7,8,2,2,22,23,2,25,26,27,28,2,3,3,32,33,3,35,36,37,38,
3,2,3,5,6,7,8,5,5,52,52,5,55,56,57,58,5,6,6,62,...
(Take out all square digits of n.)
48)
A) Smarandache Polygons:
Definition:
Let V V ...V be an n-sided polygon, n >= 3, and S a given set of m
1 2 n
elements, m >= n. In each vertex V ,1 <= i <= n, of the polygon one put
i
at most an element of S, and on each side s ,1 <= i <= n, of the polygon
i
one put e >= 0 elements from A. Let R be a given relationship of two
i
or more elements on S.
All elements of S should be put on the polygon's sides and vertexes such
that the relationship R on each side give the same result.
What connections should be among the number of sides of the polygon,
the set S (what elements and how many), and the relation R
in oredr for this problem to have solution(s) ?
B) Smarandache Polyhedrons:
I) Smarandache Polyhedrons (With Edge Points):
Similar definition generalized in a 3-dimensional space:
the points are set on the vertexes, and edges only of the polyhedron
(not on inside faces).
II) Smarandache Polyhedrons (With Face Points):
Similar definition generalized in a 3-dimensional space:
the points are set on the vertexes and inside faces of the
polyhedron (not on the edges).
III) Smarandache Polyhedrons (With Edge/Face Points):
Similar definition generalized in a 3-dimensional space:
the points are set on the vertexes, on edges, and on inside faces of
the polyhedron.
IV) Smarandache Polyhedrons (With Edge Points):
Similar definition generalized in a 3-dimensional space:
the points are set on the edges, and inside faces only of the
polyhedron (not in vertexes).
What connections should be among the number of vertexes/faces/edges of
the polyhedron, the set S (what elements and how many), and the relation
R in oredr for this problem to have solution(s) ?
[v + f = e + 2, Euler's Result, where v = number of vertexes of the
polyhedron, f = number of faces, e = number of edges.]
Examples: I)
a) For an equilateral triangle, the set N6 = {1,2,3,4,5,6}, the addition,
and on each side to have three elements exactly, and in each vertex
an element, one gets:
1
6 5 the sum on each of the three sides is 9;
2 4 3
[the minimum elements 1,2,3 are put in the vertexes]
6
1 2 the sum on each of the three sides is 12;
5 3 4
[the maximum elements 6,5,4 are put in the vertexes]
There are no other possible combinations to keep (the sum constant).
Therefore:
The Smarandache Triangular Index SGI(3) = (9, 12; 2),
which means: the minimum value, the maximum value, the total number of
combinations respectively.
b) For a square, the set N12 = {1,2,3,4,5,6, ..., 12}, the addition,
and on each side to have four elements exactly, and in each vertex
an element.
What is the Smarandache Square Index SGI(4) ?
c) For a regular pentagon, the set N20 = {1,2,3,4,5,6, ..., 20},
the addition, and on each side to have five elements exactly,
and in each vertex an element.
What is the Smarandache Pentagonal Index SGI(5) ?
d) For a regular hexagon, the set N30 = {1,2,3,4,5,6, ..., 30},
the addition, and on each side to have six elements exactly.
What is the Smarandache Hexaagonal Index SGI(6) ?
e) And generally speaking:
For an n-sided regular polygon,
the set N(n^2-n) = {1,2,3,4,5,6, ..., n^2-n},
the addition, and on each side to have n elements exactly.
What is the Smarandache n-Sided Regular Polygon Index SGI(n) ?
Exemples: II)
a) For a regular tetrahedron (4 vertexes, 4 triangular faces, 6 edges),
the set N10 = {1,2,3,...,10}, the addition, and on each edge to have 3
elements exactly (in each vertex one element) -- to get the same sum.
b) For a regular hexahedron (8 vertexes, 6 square faces, 12 edges),
the set N32 = {1,2,3,...,32}, the addition, and on each edge to have 4
elements exactly (in each vertex one element) -- to get the same sum.
c) For a regular octahedron (6 vertexes, 8 triangular faces, 12 edges),
the set N18 = {1,2,3,...,18}, the addition, and on each edge to have 3
elements exactly (in each vertex one element) -- to get the same sum.
d) For a regular dodecahedron (14 vertexes, 12 square faces, 24 edges),
the set N62 = {1,2,3,...,62}, the addition, and on each edge to have 4
elements exactly (in each vertex one element) -- to get the same sum.
Compute the Smarandache Polyhedron Index SHI(v,e,f) = (m, M; nc)
in each case,
where v = number of vertexes of the polyhedron, e = number of edges,
f = number of faces,
and m = minimum sum value, M = maximum sum value, nc = the total number
of combinations.
e) Or other versions for the previous particular polyhedrons:
- putting elements on inside faces and vertexes only (not on edges);
- or putting elements on inside faces and edges and vertexes too;
- or putting elements on inside faces and edges only (not on vertexes).
49) Smarandache Lucky Numbers:
A number is called a Smarandache Lucky Number if by a wrong operation
one gets to the true result!
For example:
64 4
---- = --- = 4,
16 1
simplifying the digit 6 of the numerator with the digit 6 of the
denominator (wrong operation!), one gets the right result 4
(because 64 divided by 16 is equal to 4).
The wrong operation should somehow be similar to a correctoperation,
see our students' common math mistakes:
__ __ ____
\/a - \/b = \/a-b , etc.
of course avoiding the trivial cases (if b = 0 or a = b, in the
previous case, for example).
Question: how many such numbers can be produced?
50) The Sequence of 3's and 1's containing many primes:
31, 331, 3331, 33331, 333331, 3333331, 33333331, 333333331, ...
51) The Sequence of 3's and 1's containing many primes:
31, 31331, 313313331, 31331333133331, 31331333133331333331,
313313331333313333313333331,
31331333133331333331333333133333331, ...
52) Generalization:
Let f, g, h, ... be n integer functions. By concatenation of all of
them one gets a composed sequence of the form:
_______________ _______________
f(1)g(1)h(1)..., ..., f(n)g(n)h(n)...,
where some function value(s) f(i) or g(i) or h(i), for i >= 1,
may be empty.
53) Recurrence of Smarandache type Function:
A function
f: N ---> N
such that if (a, b) = 1, then f(ab) = max {f(a), f(b)}.
(Definition introduced by Sabin Tabirca, England.)
For example:
a) f(n) = 1
b) g(n) = max {p, where p is prime and p divides n},
54) Smarandache Double Factorial Function:
SD(n) is the smallest integer such that SD(n)!! is divisible by n,
where m!! is the double factorial of m, i.e.
m!! = 2x4x6x...xm if m is even,
and m!! = 1x3x5x...xm if m is odd.
55) SMARANDACHE PARADOXIST NUMBERS:
There exist a bunch of "Smarandache" number sequences.
A number n is called a "Smarandache paradoxist number" if and only if
n doesn't belong to any of the Smarandache defined numbers.
Question: find the Smarandache paradoxist number sequence.
Solution (?)
If a number k is a Smarandache paradoxist number, then k doesn't belong
to any of the Smarandache defined numbers,
therefore k doesn't belong to the Smarandache paradoxist numbers too!
If a number k doesn't belong to any of the Smarandache defined numbers,
then k is a Smarandache paradoxist number,
therefore k belongs to a Smarandache defined numbers (because Smarandache
paradoxist numbers is also in the same category) --contradiction.
Dilemma: is the Smarandache paradoxist number sequence empty ??
56) NON-SMARANDACHE NUMBERS:
A number n is called a "non-Smarandache number" if and only if
n is neither a Smarandache paradoxist number nor any of the
Smarandache defined numbers.
Question: find the non-Smarandache number sequence.
Dilemma 1: is the non-Smarandache number sequence empty, too ??
Dilemma 2: is a non-Smarandache number equivalent to a Smarandache
paradoxist number ??? (this would be another paradox !! ...because
a non-Smarandache number is not a Smarandache paradoxist number).
THE PARADOX OF SMARANDACHE NUMBERS:
Any number is a Smarandache number, the non-Smarandache number too.
(This is deduced from the following paradox (see [2]):
"All is possible, the impossible too!")ิ
References:
[1] Arizona State University, Hayden Library, "The FlorentinSmarandache
papers" special collection, Tempe, AZ 85287-1006, USA,phone:
(602)965-6515 (Carol Moore librarian), email:ICCLM@ASUACAD.BITNET .
[2] Charles T. Le, "The Smarandache Class of Paradoxes", in , Bombay, India, 1995;
and in , Salinas, CA, 1993, and in , Bucharest,
No. 2, 1994.
[3] N. J. A. Sloane and S. Plouffe, "The Encyclopedia of Integer
Sequences", Academic Press, 1995;
also online, email: superseeker@research.att.com (SUPERSEEKER by
N. J. A. Sloane, S. Plouffe, B. Salvy, ATT Bell Labs,Murray Hill,
NJ 07974, USA).
[4] Henry Ibstedt, "Surfing On the Ocean of Numbers -- A Few Smarandache
Notions and Similar Topics", Erhus Univ. Press, Vail, USA, 1997.
[5] Fanel Iacobescu, "Smarandache Partition Type and Other Sequences",
, Bombay, India, Vol. 16E,
No. 2, 1997, pp. 237-40.
[6] Anthony Begay, "Smarandache Ceil Functions",
, Bombay, India, Vol. 16E,
No. 2, 1997, pp. 227-9.
[7] Helen Marimutha, "Smarandache Concatenate Type Sequences",
, Bombay, India, Vol. 16E,
No. 2, 1997, pp. 225-6.
[8] Mihaly Bencze, "Smarandache Recurrence Type Sequences",
, Bombay, India, Vol. 16E,
No. 2, 1997, pp. 231-36.
[9] Mihaly Bencze, "Smarandache Relationships and Sub-Sequences",
, Bombay, India, Vol. 17E,
No. 1, June 1998, pp. 89-95.
[10] Henry Ibstedt, "Computer Analysis of Sequences", American Research
Press, Lupton, 1998.