arXiv:0704.0875v2  [math.GM]  7 Apr 2007
The concrete theory of numbers: initial numbers
and wonderful properties of numbers repunit
Tarasov, B.V.∗
November 15, 2021
Abstract
In this work initial numbers and repunit numbers have been studied.
All numbers have been considered in a decimal notation. The problem of
simplicity of initial numbers has been studied. Interesting properties of
numbers repunit are proved: gcd(Ra, Rb) = Rgcd(a,b); Rab/(RaRb)
is an integer only if gcd(a, b) = 1, where a ≥ 1, b ≥ 1 are integers.
Dividers of numbers repunit, are researched by a degree of prime number.
Devoted to the tercentenary from the date of birth (4/15/1707)
of Leonhard Euler
1 Introduction
Let x ≥ 0, n ≥ 0 be integers. An integer N , which record consists from n
records of number x, we shall designate by
N = {x}n = x . . . x, n > 0. (1)
For n = 0 it is received {x}0 = ∅ an empty record. For example, {10}31 =
1010101, {10}01 = 1, etc.
Palindromic numbers of a kind
En,k = {1{0}k}n1, (2)
where n ≥ 0, k ≥ 0 we will name initial numbers. We will notice thatE0,k = 1
at any k ≥ 0.
Numbers repunit(see[2, 3, 4]) are natural numbers, which records consist of
units only, i.e. by definition
Rn = En−1,0, (3)
where n ≥ 1.
In decimal notation the general formula for numbers repunit is
Rn = (10
n − 1)/9, (4)
∗Tarasov, B.V. The concrete theory of numbers: initial numbers and wonderful properties
of numbers repunit. MSC 11A67+11B99. c©2007 Tarasov, B.V., independent researcher.
http://arxiv.org/abs/0704.0875v2
Tarasov, B.V. ”Initial and repunit numbers” 2
where n = 1, 2, 3, . . . .
There are known only five prime repunit for n =2,19, 23, 317, 1031.
Known problem ((Prime repunit numbers[3])). Whether exists infinite num-
ber of prime numbers repunit ?
Will we use designations further :
(a, b) = gcd(a, b) the greatest common divider of integers a > 0, b > 0.
p, q odd prime numbers.
If it is not stipulated specially, the integer positive numbers are considered.
2 Initial numbers
Let’s consider the trivial properties of initial numbers.
Theorem 1. Following trivial statements are fair :
(1) General formula of initial numbers is
En,k =
R(k+1)(n+1)
10(k+1)(n+1) − 1
10k+1 − 1
. (5)
(2) For k ≥ 0, n ≥ m ≥ 1 if n + 1 ≡ 0(mod (m + 1)),
then (En,k, Em,k) = Em,k.
(3) For k ≥ 0, n > m ≥ 1 if integer s ≥ 1, exists such
that n + 1 ≡ 0(mod (s + 1)), m + 1 ≡ 0(mod (s + 1)), then
(En,k, Em,k) ≥ Es,k > 1.
(4) For k ≥ 0, n > m ≥ 1 (En,k, Em,k) = 1 when and only then,
(n + 1,m + 1) = 1.
Proof. 1) Properties (1)—(3) are obvious.
2) The Proof of property (4). Necessity. Let
(En,k, Em,k) = 1 and (n + 1,m + 1) = s > 1, s − 1 ≥ 1. From property
(3) of the theorem follows that (En,k, Em,k) ≥ Es−1,k = {1{0}k}s−11 > 1.
Appears the contradiction .
Sufficiency of property (4). Let (n+1,m+1) = 1, then will be integers
a > 0, b > 0, such that either a(n + 1) = b(m + 1) + 1
or b(m + 1) = a(n + 1) + 1. Let’s assume, that (En,k, Em,k) = d > 1.
a) Let a(n + 1) = b(m + 1) + 1, then Eb(m+1),k = Ea(n+1)−1,k =
(10a(n+1)(k+1) − 1)/(10k+1 − 1) ≡ 0(modEn,k) ≡ 0(modd).
On the other hand Eb(m+1),k = (10
(k+1){b(m+1)+1}−1)/(10k+1−1) =
((10b(m+1)(k+1) − 1)/(10k+1 − 1)) · 10k+1 + 1 ≡
≡ 1(modEm,k) ≡ 1(modd). Appears the contradiction.
b) Let b(m + 1) = a(n + 1) + 1, then Ea(n+1),k = Eb(m+1)−1,k =
(10b(m+1)(k+1) − 1)/(10k+1 − 1) ≡ 0(modEm,k) ≡ 0(modd).
On the other hand Ea(n+1),k = (10
(k+1){a(n+1)+1} −1)/(10k+1−1) =
((10a(n+1)(k+1) − 1)/(10k+1 − 1)) · 10k+1 + 1 ≡
≡ 1(modEn,k) ≡ 1(modd). Have received the contradiction.
3 Numbers repunit
Let’s consider trivial properties of numbers repunit.
Tarasov, B.V. ”Initial and repunit numbers” 3
Theorem 2. Following trivial statements are fair :
(1) The number Rn is prime only if n number is prime.
(2) If p > 3 all prime dividers of number Rp look like 1+2px where x ≥ 1
is integer.
(3) (Ra, Rb) = 1 if and only if (a, b) = 1.
Proof. Property (1) of theorem is proved in ([2, 3]), property (2) is proved in
([1]), as exercise. Property (3) is the corollary of the theorem 1.
Theorem 3. (Ra, Rb) = R(a,b), where a ≥ 1, b ≥ 1 are integers.
Proof. Validity of the theorem for (a, b) = 1 follows from property (3) of
theorem2. Let (a, b) = d > 1, where a = a1d, b = b1d, (a1, b1) = 1. Let’s
consider equations
Ra = Rd · {10
d(a1−1) + . . . + 10d + 1},
Rb = Rd · {10
d(b1−1) + . . . + 10d + 1}.
A = 10d(a1−1) + . . . + 10d + 1,
B = 10d(b1−1) + . . . + 10d + 1.
Let’s assume, that (A,B) > 1, and q is a prime odd number such that
A ≡ 0(modq), B ≡ 0(modq). (6)
If q = 3, then 10t ≡ 1(modq) for any integer t ≥ 1. Then from (6) it
follows that a1 ≡ b1 ≡ 0(modq). Have received the contradiction.
Thus, q > 3. Then there exists an index dmin, to which the number 10
belongs on the module q.
(10d)dmin ≡ 1(modq),
where dmin ≥ 1.
If dmin = 1, then it follows from (6) that a1 ≡ b1 ≡ 0(modq). Have
received the contradiction. Hence dmin > 1. As Ra ≡ Rb ≡ 0(modq),
then (10d)a1 ≡ 1(modq) and (10d)b1 ≡ 1(modq).
Then a1 ≡ b1 ≡ 0(moddmin). Have received the contradiction.
Theorem 4. Let p > 3 be a prime number, k ≥ t ≥ 1, t ≥ s ≥ 1 integer
numbers. Then
gcd(Rpk/Rpt , Rps) = 1. (7)
Proof. Let’s consider expression
A = Rpk/Rpt = (10
k−t−1 + (10p
k−t−2 + . . . + 10p
If (A,Rps) > 1, then the prime number q exists such that
A ≡ 0(modq) Rps ≡ 0(modq). Hence 10
≡ 1(modq), then
A ≡ pk−t ≡ 0(modq), p = q = 3. Have received the contradiction,
because p > 3.
Tarasov, B.V. ”Initial and repunit numbers” 4
Theorem 5. Let a ≥ 1, b ≥ 1 are integers, then the following statements are
true :
(1) If (a, b) = 1, then
gcd(Rab, RaRb) = RaRb. (8)
(2) If (a, b) > 1, then
RaRb/R(a,b) ≤ gcd(Rab, RaRb) < RaRb. (9)
Proof. 1) Let (a, b) = 1, then (Ra, Rb) = R(a,b) = 1,
Rab = RaX = RbY , X = cRb, where c ≥ 1 is integer. Rab = cRaRb.
2) Let (a, b) = d > 1, a = a1d, b = b1d, (a1, b1) = 1, a1 ≥ 1, b1 ≥ 1.
As gcd(Ra, Rb) = R(a,b), we receive equality
Ra = R(a,b)X,Rb = R(a,b)Y, (10)
where (X,Y ) = 1.
Further, Rab = RaA = RbB = XAR(a,b) = Y BR(a,b), XA = Y B,
A = Y z, B = Xz, z ≥ 1 is integer. Then Rab = XY R(a,b)z,
Rab = zRaRb/R(a,b). We have proved, that
RaRb/R(a,b) ≤ gcd(Rab, RaRb).
Let’s assume, that gcd(Rab, RaRb) = RaRb, then Rab = zRaRb, where
z ≥ 1 is integer. Let’s consider equalities
Rab = RaA = RbB,
where
A = 10a(b−1) + 10a(b−2) + . . . + 10a + 1,
B = 10b(a−1) + 10b(a−1) + . . . + 10b + 1.
Since A = Rbz, B = Raz, 10
a ≡ 1(modR(a,b)),
10b ≡ 1(modR(a,b)), then A ≡ B ≡ 0(modR(a,b)), hence
a ≡ b ≡ 0(modR(a,b)).
Thus, comparison (a, b) ≡ 0(modR(a,b)) or
d ≡ 0(modRd) is fair, that contradicts an obvious inequality
(10x − 1)/9 > x, (11)
where x > 1 is real.
({⋆} The Important corollary of the theorem 5).
Number Rab/(RaRb) is integer when and only when (a, b) = 1, where a ≥ 1,
b ≥ 1 are integers.
Let’s quote some trivial statements for numbers repunit.
Lemma 1. If a = 3nb, (b, 3) = 1, then
Ra ≡ 0(mod 3
n), butRa 6≡ 0(mod 3
(n+1)). (12)
Tarasov, B.V. ”Initial and repunit numbers” 5
Proof. If n = 1, then Ra = R3B, where B = 10
3(b−1) + . . . + 103 + 1,
R3 ≡ 0(mod 3), B ≡ b 6≡ 0(mod 3). Thus,
Ra ≡ 0(mod 3), but Ra 6≡ 0(mod 3
Let comparisons (12) be proved for n ≤ k− 1. We shall consider a = 3kb,
(b, 3) = 1. Then Ra = R3k−1bA, where A = 10
3k−1b2 + 103
b + 1.
R3k−1b ≡ 0(mod 3
k−1), but R3k−1b 6≡ 0(mod 3
k), A ≡ 0(mod 3), but
A 6≡ 0(mod 32).
Lemma 2. If n ≥ 0 is integer, then
rn = 10
11n + 1 ≡ 0(mod 11n+1), but rn 6≡ 0(mod 11
n+2). (13)
Proof. r0 = 11 ≡ 0(mod 11), but r0 = 11 6≡ 0(mod 11
r1 = 10
11 + 1 ≡ 0(mod 112), but r1 6≡ 0(mod 11
Let’s make the inductive assumption, that formulas (13) are proved for
n ≤ k − 1, where k − 1 ≥ 1, k ≥ 2. Let n = k, then
rk = 10
11k + 1 = (1011
)11 + 1 = rk−1A, where
A = 1011
k−110 − 1011
k−19 + 1011
k−18 − 1011
k−17 + 1011
k−16−
− 1011
k−15 + 1011
k−14 − 1011
k−13 + 1011
k−12 − 1011
+ 1. (14)
Since, due to the inductive assumption 1011
≡ −1(mod 11k), where
k ≥ 2, then A ≡ 11(mod 11k). Then A ≡ 0(mod 11), but
A 6≡ 0(mod 112). Thus, we receive, that rk ≡ 0(mod 11
k+1), but
rk 6≡ 0(mod 11
k+2).
Lemma 3. For an integer a ≥ 1, the following statements are true :
(1) If a is odd, then Ra 6≡ 0(mod 11).
(2) If a = 2(11n)b, (b, 11) = 1, then
Ra ≡ 0(mod 11
n+1), butRa 6≡ 0(mod 11
n+2). (15)
Proof. If a is odd, then Ra ≡ 1(mod 11). If a = 2(11
n)b, (b, 11) = 1,
then Ra = ((10
2(11)n)b − 1)/9 = R11n · rn · A, where rn = 10
11n + 1,
A = 102(11
n)(b−1) + . . . + 102(11
n) + 1. R11n 6≡ 0(mod 11),
A ≡ b 6≡ 0(mod 11). Then validity of the statement (2) of lemma 3 follows
from lemma 2.
({⋆} The assumption: the general formula for gcd(Rab, RaRb)).
If a ≥ 1, b ≥ 1 are integers, d = (a, b), where d = 3L · 11S · c, (c, 3) = 1,
(c, 11) = 1, L ≥ 0, S ≥ 0, then equalities are true :
— if c is an odd number, then
gcd(Rab, RaRb) = ((RaRb)/R(a,b)) · 3
L, (16)
— if c is an even number, then
gcd(Rab, RaRb) = ((RaRb)/R(a,b)) · 3
L · 11S. (17)
Let’s give another two obvious statements in which divisors of numbers
repunit are studied, as degrees of prime number.
Tarasov, B.V. ”Initial and repunit numbers” 6
Lemma 4. If p, q are prime numbers and Rp ≡ 0(modq), but
Rp 6≡ 0(modq
2), then statements are true :
(1) For any integer r, 0 < r < q, Rpr 6≡ 0(modq
(2) For any integer n, n ≥ 1, Rpn 6≡ 0(modq
Proof. 1) Rpr = Rp · R̂pr, where R̂pr = 10
p(r−1) + 10p(r−2) +
+ . . . + 10p + 1. If Rpr ≡ 0(modq
2), then R̂pr ≡ 0(modq),
r ≡ 0(modq). Have received the contradiction.
2) If n > 1 found such that Rpn ≡ 0(modq
2), then from (7) follows
(Rpn/Rp, Rp) = 1. Have received the contradiction.
Lemma 5. If p, q are prime numbers and Rp ≡ 0(modq), then
Rpqn ≡ 0(modq
n+1).
Proof. Since Rpq = Rp · R̂pq, where R̂pq = 10
p(q−1) +
+ 10p(q−2) + . . . + 10p + 1, then R̂pq ≡ 0(modq), Rpq ≡ 0(modq
Let’s assume that Rpqn−1 ≡ 0(modq
n). Then
Rpqn = Rpqn−1·q = Rpqn−1 · R̂pqn−1·q, where
R̂pqn−1·q = 10
n−1·(q−1)+10pq
n−1·(q−2)+ . . .+10pq
+1 ≡ 0(modq),
Rpqn ≡ 0(modq
n+1).
4 Problem of simplicity of initial numbers
Let’s consider the problem of simplicity of initial numbers En,k, where
k ≥ 0, n ≥ 0.
If k = 0, then En,0 = Rn+1. Thus, simplicity of numbers En,0 – is known
problem of prime numbers repunit Rp, where p is prime number.
If n = 1, then E1,k = 1{0}k1 = 10
k+1 + 1. As number E1,k can be
prime only when k + 1 = 2m, m ≥ 0 is integer, then we come to the known
problem of simplicity of the generalized Fermat numbers fm(a) = a
2m + 1
for a = 10. Generalized Fermat numbers nave been define by Ribenboim [5] in
1996, as numbers of the form fn(a) = a
2n + 1, where a > 2 is even.
The generalized Fermat numbers fn(10) = 10
2n + 1 for n ≤ 14 are prime
only if n = 0, 1. f0(10) = 11, f1(10) = 101.
Theorem 6. Let n > 1, k > 0. If any of conditions
(1) n number is odd,
(2) k number is odd,
(3) n + 1 ≡ 0(mod 3),
(4) (n + 1, k + 1) = 1,
is true, then number En,k is compound.
Proof. 1) n + 1 = 2t, t > 1. Then En,k = Et−1,k · (10
t(k+1) + 1), where
t > 1, t − 1 ≥ 1. As Et−1,k > 1, then En,k is compound number.
2) Let k be an odd number. Due to the proved condition (1) we count that
number (n + 1) is odd. k + 1 = 2t ≥ 2, t ≥ 1. Further,
En,k = En,t−1 · ((10
(n+1)t + 1)/(10t + 1)),
where n > 1, t− 1 ≥ 0, En,t−1 > 1, number (10
(n+1)t +1)/(10t +1) > 1
is integer.
Tarasov, B.V. ”Initial and repunit numbers” 7
3) If n + 1 ≡ 0(mod 3), then En,k ≡ 0(mod 3), En,k > 11.
4) Let n > 1, k ≥ 1, (n + 1, k + 1) = 1, then
En,k = R(n+1)(k+1)/R(k+1) = R(n+1) · (R(n+1)(k+1)/(Rk+1 · Rn+1)).
Due to the theorem 5 number z = R(n+1)(k+1)/(Rk+1 · Rn+1) is integer.
Further,
z > (10(n+1)(k+1)−1)/(10n+k+2) = 10nk−1−1/(10n+k+2), nk−1 ≥ 1,
thus, z > 1.
Question of simplicity of initial numbers under conditions, when
(n + 1, k + 1) > 1, (n + 1) number is odd, (k + 1) number is odd,
n + 1 6≡ 0(mod 3), remains open.
In particular, it is interesting to considerate numbers Ep−1,p−1 = Rp2/Rp,
where p is prime number. For p < 100 numbers Ep−1,p−1 are compound.
5 The open problems of numbers repunit
The known problem of numbers repunit remains open.
Problem 1 ((Prime repunit numbers[3])). Whether there exists infinite number
of prime numbers Rp, p–prime number ?
Problem 2. Whether all numbers Rp, p–prime number, are numbers free from
squares ?
The author has checked up for p < 97, that numbers Rp are free from
squares. Another following open questions are interesting :
Problem 3. If number Rp is free from squares, where p > 3 is prime number,
whether will number n, be found such what number Rpn contains a square ?
Problem 4. p is prime number, whether there are simple numbers of a kind
Ep−1,p−1 = Rp2/Rp ?
The author has checked up to p ≤ 127, that numbers Ep−1,p−1 is com-
pound. It is known, that Rp divide by number (2p + 1) for prime numbers
p = 41, 53, Rp divide by number (4p+1) for prime numbers p = 13, 43, 79.
There appears a question :
Problem 5. Whether there is infinite number of prime numbers p, such that
Rp divide by number (2p + 1) or is number (4p + 1) ?
(The remark). If the number p > 5 Sophie Germain prime (i.e. number
2p+1 is prime too), then either Rp or R
= (10p +1)/11 divide by number
(2p + 1).
6 The conclusion
Leonhard Euler, professor of the Russian Academy of sciences since 1731,
has paid mathematics forever ! Euler’s invisible hand directs the develop-
ment of concrete mathematics for more than 200 years.
Euler’s titanic work which has opened a way to freedom to mathematical
community, admires. The pleasure caused by Euler’s works warms hearts.
Tarasov, B.V. ”Initial and repunit numbers” 8
References
[1] Vinogradov I.M. Osnovy teorii chisel. -M. :Nauka, 1981.
[2] Ronald L.Graham,Donald E.Knuth,Oren Patashnik,
Concrete Mathematics : A Foundation for Computer Science, 2nd edition
(Reading,Massachusetts: Addison-Wesley), 1994.
[3] Weisstein, Eric W. ”Repunit.” From MathWorld–A Wolfram Web Resource.
—http://mathworld.wolfram.com/Repunit.html/.
c©1999—2007 Wolfram Research, Inc.
[4] The Prime Clossary repunit.
—http://primes.utm.edu/glossary/page.php?sort=Repunit/.
[5] Ribenboim, P. ”Fermat Numbers” and ”Numbers k × 2n ± 1.” 2.6 and 5.7
in The New Book of Prime Number Records. New York: Springer-Verlag,
pp. 83-90 and 355-360, 1996.
———————————————————————
Institute of Thermophysics, Siberian Branch of RAS
Lavrentyev Ave., 1, Novosibirsk, 630090, Russia
E-mail: tarasov@itp.nsc.ru
———————————————————————
Independent researcher,
E-mail: tarasov-b@mail.ru
———————————————————————

  In this work initial numbers and repunit numbers have been studied. All
numbers have been considered in a decimal notation. The problem of simplicity
of initial numbers has been studied. Interesting properties of numbers repunit
are proved:
  $gcd(R_a, R_b) = R_{gcd(a,b)}$;
  $R_{ab}/(R_aR_b)$ is an integer only if $gcd(a,b) = 1$, where $a\geq1$,
$b\geq1$ are integers. Dividers of numbers repunit, are researched by a degree
of prime number.

Introduction
Let x ≥ 0, n ≥ 0 be integers. An integer N , which record consists from n
records of number x, we shall designate by
N = {x}n = x . . . x, n > 0. (1)
For n = 0 it is received {x}0 = ∅ an empty record. For example, {10}31 =
1010101, {10}01 = 1, etc.
Palindromic numbers of a kind
En,k = {1{0}k}n1, (2)
where n ≥ 0, k ≥ 0 we will name initial numbers. We will notice thatE0,k = 1
at any k ≥ 0.
Numbers repunit(see[2, 3, 4]) are natural numbers, which records consist of
units only, i.e. by definition
Rn = En−1,0, (3)
where n ≥ 1.
In decimal notation the general formula for numbers repunit is
Rn = (10
n − 1)/9, (4)
∗Tarasov, B.V. The concrete theory of numbers: initial numbers and wonderful properties
of numbers repunit. MSC 11A67+11B99. c©2007 Tarasov, B.V., independent researcher.
http://arxiv.org/abs/0704.0875v2
Tarasov, B.V. ”Initial and repunit numbers” 2
where n = 1, 2, 3, . . . .
There are known only five prime repunit for n =2,19, 23, 317, 1031.
Known problem ((Prime repunit numbers[3])). Whether exists infinite num-
ber of prime numbers repunit ?
Will we use designations further :
(a, b) = gcd(a, b) the greatest common divider of integers a > 0, b > 0.
p, q odd prime numbers.
If it is not stipulated specially, the integer positive numbers are considered.
2 Initial numbers
Let’s consider the trivial properties of initial numbers.
Theorem 1. Following trivial statements are fair :
(1) General formula of initial numbers is
En,k =
R(k+1)(n+1)
10(k+1)(n+1) − 1
10k+1 − 1
. (5)
(2) For k ≥ 0, n ≥ m ≥ 1 if n + 1 ≡ 0(mod (m + 1)),
then (En,k, Em,k) = Em,k.
(3) For k ≥ 0, n > m ≥ 1 if integer s ≥ 1, exists such
that n + 1 ≡ 0(mod (s + 1)), m + 1 ≡ 0(mod (s + 1)), then
(En,k, Em,k) ≥ Es,k > 1.
(4) For k ≥ 0, n > m ≥ 1 (En,k, Em,k) = 1 when and only then,
(n + 1,m + 1) = 1.
Proof. 1) Properties (1)—(3) are obvious.
2) The Proof of property (4). Necessity. Let
(En,k, Em,k) = 1 and (n + 1,m + 1) = s > 1, s − 1 ≥ 1. From property
(3) of the theorem follows that (En,k, Em,k) ≥ Es−1,k = {1{0}k}s−11 > 1.
Appears the contradiction .
Sufficiency of property (4). Let (n+1,m+1) = 1, then will be integers
a > 0, b > 0, such that either a(n + 1) = b(m + 1) + 1
or b(m + 1) = a(n + 1) + 1. Let’s assume, that (En,k, Em,k) = d > 1.
a) Let a(n + 1) = b(m + 1) + 1, then Eb(m+1),k = Ea(n+1)−1,k =
(10a(n+1)(k+1) − 1)/(10k+1 − 1) ≡ 0(modEn,k) ≡ 0(modd).
On the other hand Eb(m+1),k = (10
(k+1){b(m+1)+1}−1)/(10k+1−1) =
((10b(m+1)(k+1) − 1)/(10k+1 − 1)) · 10k+1 + 1 ≡
≡ 1(modEm,k) ≡ 1(modd). Appears the contradiction.
b) Let b(m + 1) = a(n + 1) + 1, then Ea(n+1),k = Eb(m+1)−1,k =
(10b(m+1)(k+1) − 1)/(10k+1 − 1) ≡ 0(modEm,k) ≡ 0(modd).
On the other hand Ea(n+1),k = (10
(k+1){a(n+1)+1} −1)/(10k+1−1) =
((10a(n+1)(k+1) − 1)/(10k+1 − 1)) · 10k+1 + 1 ≡
≡ 1(modEn,k) ≡ 1(modd). Have received the contradiction.
3 Numbers repunit
Let’s consider trivial properties of numbers repunit.
Tarasov, B.V. ”Initial and repunit numbers” 3
Theorem 2. Following trivial statements are fair :
(1) The number Rn is prime only if n number is prime.
(2) If p > 3 all prime dividers of number Rp look like 1+2px where x ≥ 1
is integer.
(3) (Ra, Rb) = 1 if and only if (a, b) = 1.
Proof. Property (1) of theorem is proved in ([2, 3]), property (2) is proved in
([1]), as exercise. Property (3) is the corollary of the theorem 1.
Theorem 3. (Ra, Rb) = R(a,b), where a ≥ 1, b ≥ 1 are integers.
Proof. Validity of the theorem for (a, b) = 1 follows from property (3) of
theorem2. Let (a, b) = d > 1, where a = a1d, b = b1d, (a1, b1) = 1. Let’s
consider equations
Ra = Rd · {10
d(a1−1) + . . . + 10d + 1},
Rb = Rd · {10
d(b1−1) + . . . + 10d + 1}.
A = 10d(a1−1) + . . . + 10d + 1,
B = 10d(b1−1) + . . . + 10d + 1.
Let’s assume, that (A,B) > 1, and q is a prime odd number such that
A ≡ 0(modq), B ≡ 0(modq). (6)
If q = 3, then 10t ≡ 1(modq) for any integer t ≥ 1. Then from (6) it
follows that a1 ≡ b1 ≡ 0(modq). Have received the contradiction.
Thus, q > 3. Then there exists an index dmin, to which the number 10
belongs on the module q.
(10d)dmin ≡ 1(modq),
where dmin ≥ 1.
If dmin = 1, then it follows from (6) that a1 ≡ b1 ≡ 0(modq). Have
received the contradiction. Hence dmin > 1. As Ra ≡ Rb ≡ 0(modq),
then (10d)a1 ≡ 1(modq) and (10d)b1 ≡ 1(modq).
Then a1 ≡ b1 ≡ 0(moddmin). Have received the contradiction.
Theorem 4. Let p > 3 be a prime number, k ≥ t ≥ 1, t ≥ s ≥ 1 integer
numbers. Then
gcd(Rpk/Rpt , Rps) = 1. (7)
Proof. Let’s consider expression
A = Rpk/Rpt = (10
k−t−1 + (10p
k−t−2 + . . . + 10p
If (A,Rps) > 1, then the prime number q exists such that
A ≡ 0(modq) Rps ≡ 0(modq). Hence 10
≡ 1(modq), then
A ≡ pk−t ≡ 0(modq), p = q = 3. Have received the contradiction,
because p > 3.
Tarasov, B.V. ”Initial and repunit numbers” 4
Theorem 5. Let a ≥ 1, b ≥ 1 are integers, then the following statements are
true :
(1) If (a, b) = 1, then
gcd(Rab, RaRb) = RaRb. (8)
(2) If (a, b) > 1, then
RaRb/R(a,b) ≤ gcd(Rab, RaRb) < RaRb. (9)
Proof. 1) Let (a, b) = 1, then (Ra, Rb) = R(a,b) = 1,
Rab = RaX = RbY , X = cRb, where c ≥ 1 is integer. Rab = cRaRb.
2) Let (a, b) = d > 1, a = a1d, b = b1d, (a1, b1) = 1, a1 ≥ 1, b1 ≥ 1.
As gcd(Ra, Rb) = R(a,b), we receive equality
Ra = R(a,b)X,Rb = R(a,b)Y, (10)
where (X,Y ) = 1.
Further, Rab = RaA = RbB = XAR(a,b) = Y BR(a,b), XA = Y B,
A = Y z, B = Xz, z ≥ 1 is integer. Then Rab = XY R(a,b)z,
Rab = zRaRb/R(a,b). We have proved, that
RaRb/R(a,b) ≤ gcd(Rab, RaRb).
Let’s assume, that gcd(Rab, RaRb) = RaRb, then Rab = zRaRb, where
z ≥ 1 is integer. Let’s consider equalities
Rab = RaA = RbB,
where
A = 10a(b−1) + 10a(b−2) + . . . + 10a + 1,
B = 10b(a−1) + 10b(a−1) + . . . + 10b + 1.
Since A = Rbz, B = Raz, 10
a ≡ 1(modR(a,b)),
10b ≡ 1(modR(a,b)), then A ≡ B ≡ 0(modR(a,b)), hence
a ≡ b ≡ 0(modR(a,b)).
Thus, comparison (a, b) ≡ 0(modR(a,b)) or
d ≡ 0(modRd) is fair, that contradicts an obvious inequality
(10x − 1)/9 > x, (11)
where x > 1 is real.
({⋆} The Important corollary of the theorem 5).
Number Rab/(RaRb) is integer when and only when (a, b) = 1, where a ≥ 1,
b ≥ 1 are integers.
Let’s quote some trivial statements for numbers repunit.
Lemma 1. If a = 3nb, (b, 3) = 1, then
Ra ≡ 0(mod 3
n), butRa 6≡ 0(mod 3
(n+1)). (12)
Tarasov, B.V. ”Initial and repunit numbers” 5
Proof. If n = 1, then Ra = R3B, where B = 10
3(b−1) + . . . + 103 + 1,
R3 ≡ 0(mod 3), B ≡ b 6≡ 0(mod 3). Thus,
Ra ≡ 0(mod 3), but Ra 6≡ 0(mod 3
Let comparisons (12) be proved for n ≤ k− 1. We shall consider a = 3kb,
(b, 3) = 1. Then Ra = R3k−1bA, where A = 10
3k−1b2 + 103
b + 1.
R3k−1b ≡ 0(mod 3
k−1), but R3k−1b 6≡ 0(mod 3
k), A ≡ 0(mod 3), but
A 6≡ 0(mod 32).
Lemma 2. If n ≥ 0 is integer, then
rn = 10
11n + 1 ≡ 0(mod 11n+1), but rn 6≡ 0(mod 11
n+2). (13)
Proof. r0 = 11 ≡ 0(mod 11), but r0 = 11 6≡ 0(mod 11
r1 = 10
11 + 1 ≡ 0(mod 112), but r1 6≡ 0(mod 11
Let’s make the inductive assumption, that formulas (13) are proved for
n ≤ k − 1, where k − 1 ≥ 1, k ≥ 2. Let n = k, then
rk = 10
11k + 1 = (1011
)11 + 1 = rk−1A, where
A = 1011
k−110 − 1011
k−19 + 1011
k−18 − 1011
k−17 + 1011
k−16−
− 1011
k−15 + 1011
k−14 − 1011
k−13 + 1011
k−12 − 1011
+ 1. (14)
Since, due to the inductive assumption 1011
≡ −1(mod 11k), where
k ≥ 2, then A ≡ 11(mod 11k). Then A ≡ 0(mod 11), but
A 6≡ 0(mod 112). Thus, we receive, that rk ≡ 0(mod 11
k+1), but
rk 6≡ 0(mod 11
k+2).
Lemma 3. For an integer a ≥ 1, the following statements are true :
(1) If a is odd, then Ra 6≡ 0(mod 11).
(2) If a = 2(11n)b, (b, 11) = 1, then
Ra ≡ 0(mod 11
n+1), butRa 6≡ 0(mod 11
n+2). (15)
Proof. If a is odd, then Ra ≡ 1(mod 11). If a = 2(11
n)b, (b, 11) = 1,
then Ra = ((10
2(11)n)b − 1)/9 = R11n · rn · A, where rn = 10
11n + 1,
A = 102(11
n)(b−1) + . . . + 102(11
n) + 1. R11n 6≡ 0(mod 11),
A ≡ b 6≡ 0(mod 11). Then validity of the statement (2) of lemma 3 follows
from lemma 2.
({⋆} The assumption: the general formula for gcd(Rab, RaRb)).
If a ≥ 1, b ≥ 1 are integers, d = (a, b), where d = 3L · 11S · c, (c, 3) = 1,
(c, 11) = 1, L ≥ 0, S ≥ 0, then equalities are true :
— if c is an odd number, then
gcd(Rab, RaRb) = ((RaRb)/R(a,b)) · 3
L, (16)
— if c is an even number, then
gcd(Rab, RaRb) = ((RaRb)/R(a,b)) · 3
L · 11S. (17)
Let’s give another two obvious statements in which divisors of numbers
repunit are studied, as degrees of prime number.
Tarasov, B.V. ”Initial and repunit numbers” 6
Lemma 4. If p, q are prime numbers and Rp ≡ 0(modq), but
Rp 6≡ 0(modq
2), then statements are true :
(1) For any integer r, 0 < r < q, Rpr 6≡ 0(modq
(2) For any integer n, n ≥ 1, Rpn 6≡ 0(modq
Proof. 1) Rpr = Rp · R̂pr, where R̂pr = 10
p(r−1) + 10p(r−2) +
+ . . . + 10p + 1. If Rpr ≡ 0(modq
2), then R̂pr ≡ 0(modq),
r ≡ 0(modq). Have received the contradiction.
2) If n > 1 found such that Rpn ≡ 0(modq
2), then from (7) follows
(Rpn/Rp, Rp) = 1. Have received the contradiction.
Lemma 5. If p, q are prime numbers and Rp ≡ 0(modq), then
Rpqn ≡ 0(modq
n+1).
Proof. Since Rpq = Rp · R̂pq, where R̂pq = 10
p(q−1) +
+ 10p(q−2) + . . . + 10p + 1, then R̂pq ≡ 0(modq), Rpq ≡ 0(modq
Let’s assume that Rpqn−1 ≡ 0(modq
n). Then
Rpqn = Rpqn−1·q = Rpqn−1 · R̂pqn−1·q, where
R̂pqn−1·q = 10
n−1·(q−1)+10pq
n−1·(q−2)+ . . .+10pq
+1 ≡ 0(modq),
Rpqn ≡ 0(modq
n+1).
4 Problem of simplicity of initial numbers
Let’s consider the problem of simplicity of initial numbers En,k, where
k ≥ 0, n ≥ 0.
If k = 0, then En,0 = Rn+1. Thus, simplicity of numbers En,0 – is known
problem of prime numbers repunit Rp, where p is prime number.
If n = 1, then E1,k = 1{0}k1 = 10
k+1 + 1. As number E1,k can be
prime only when k + 1 = 2m, m ≥ 0 is integer, then we come to the known
problem of simplicity of the generalized Fermat numbers fm(a) = a
2m + 1
for a = 10. Generalized Fermat numbers nave been define by Ribenboim [5] in
1996, as numbers of the form fn(a) = a
2n + 1, where a > 2 is even.
The generalized Fermat numbers fn(10) = 10
2n + 1 for n ≤ 14 are prime
only if n = 0, 1. f0(10) = 11, f1(10) = 101.
Theorem 6. Let n > 1, k > 0. If any of conditions
(1) n number is odd,
(2) k number is odd,
(3) n + 1 ≡ 0(mod 3),
(4) (n + 1, k + 1) = 1,
is true, then number En,k is compound.
Proof. 1) n + 1 = 2t, t > 1. Then En,k = Et−1,k · (10
t(k+1) + 1), where
t > 1, t − 1 ≥ 1. As Et−1,k > 1, then En,k is compound number.
2) Let k be an odd number. Due to the proved condition (1) we count that
number (n + 1) is odd. k + 1 = 2t ≥ 2, t ≥ 1. Further,
En,k = En,t−1 · ((10
(n+1)t + 1)/(10t + 1)),
where n > 1, t− 1 ≥ 0, En,t−1 > 1, number (10
(n+1)t +1)/(10t +1) > 1
is integer.
Tarasov, B.V. ”Initial and repunit numbers” 7
3) If n + 1 ≡ 0(mod 3), then En,k ≡ 0(mod 3), En,k > 11.
4) Let n > 1, k ≥ 1, (n + 1, k + 1) = 1, then
En,k = R(n+1)(k+1)/R(k+1) = R(n+1) · (R(n+1)(k+1)/(Rk+1 · Rn+1)).
Due to the theorem 5 number z = R(n+1)(k+1)/(Rk+1 · Rn+1) is integer.
Further,
z > (10(n+1)(k+1)−1)/(10n+k+2) = 10nk−1−1/(10n+k+2), nk−1 ≥ 1,
thus, z > 1.
Question of simplicity of initial numbers under conditions, when
(n + 1, k + 1) > 1, (n + 1) number is odd, (k + 1) number is odd,
n + 1 6≡ 0(mod 3), remains open.
In particular, it is interesting to considerate numbers Ep−1,p−1 = Rp2/Rp,
where p is prime number. For p < 100 numbers Ep−1,p−1 are compound.
5 The open problems of numbers repunit
The known problem of numbers repunit remains open.
Problem 1 ((Prime repunit numbers[3])). Whether there exists infinite number
of prime numbers Rp, p–prime number ?
Problem 2. Whether all numbers Rp, p–prime number, are numbers free from
squares ?
The author has checked up for p < 97, that numbers Rp are free from
squares. Another following open questions are interesting :
Problem 3. If number Rp is free from squares, where p > 3 is prime number,
whether will number n, be found such what number Rpn contains a square ?
Problem 4. p is prime number, whether there are simple numbers of a kind
Ep−1,p−1 = Rp2/Rp ?
The author has checked up to p ≤ 127, that numbers Ep−1,p−1 is com-
pound. It is known, that Rp divide by number (2p + 1) for prime numbers
p = 41, 53, Rp divide by number (4p+1) for prime numbers p = 13, 43, 79.
There appears a question :
Problem 5. Whether there is infinite number of prime numbers p, such that
Rp divide by number (2p + 1) or is number (4p + 1) ?
(The remark). If the number p > 5 Sophie Germain prime (i.e. number
2p+1 is prime too), then either Rp or R
= (10p +1)/11 divide by number
(2p + 1).
6 The conclusion
Leonhard Euler, professor of the Russian Academy of sciences since 1731,
has paid mathematics forever ! Euler’s invisible hand directs the develop-
ment of concrete mathematics for more than 200 years.
Euler’s titanic work which has opened a way to freedom to mathematical
community, admires. The pleasure caused by Euler’s works warms hearts.
Tarasov, B.V. ”Initial and repunit numbers” 8
References
[1] Vinogradov I.M. Osnovy teorii chisel. -M. :Nauka, 1981.
[2] Ronald L.Graham,Donald E.Knuth,Oren Patashnik,
Concrete Mathematics : A Foundation for Computer Science, 2nd edition
(Reading,Massachusetts: Addison-Wesley), 1994.
[3] Weisstein, Eric W. ”Repunit.” From MathWorld–A Wolfram Web Resource.
—http://mathworld.wolfram.com/Repunit.html/.
c©1999—2007 Wolfram Research, Inc.
[4] The Prime Clossary repunit.
—http://primes.utm.edu/glossary/page.php?sort=Repunit/.
[5] Ribenboim, P. ”Fermat Numbers” and ”Numbers k × 2n ± 1.” 2.6 and 5.7
in The New Book of Prime Number Records. New York: Springer-Verlag,
pp. 83-90 and 355-360, 1996.
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Institute of Thermophysics, Siberian Branch of RAS
Lavrentyev Ave., 1, Novosibirsk, 630090, Russia
E-mail: tarasov@itp.nsc.ru
———————————————————————
Independent researcher,
E-mail: tarasov-b@mail.ru
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