Does there exist prime number of the form $1010101010101....$ after the trivial one $101$?

Question

I have checked numbers of the form $101010101...$ up to $1(01)_{2500}$ and the only prime I found is 101. I found that numbers of such form are quite rich in number of distinct prime factors. And $1(01)_{18}$ is the only semiprime I found So far(!!). Are there anymore primes of such form ?

Answer 1

No, there is no other such prime. The number of this form with $n$ 1's is equal to $a_n = \frac{10^{2n}-1}{99}$. Now:

If $n$ is even, then $a_n$ is divisible by 101, since $10^{2n} = 100^n \equiv (-1)^n = 1 \pmod{101}$ and 99 is relatively prime to 101. So if $n>2$ also, then $a_n$ is composite.

If $n$ is odd, then $a_n$ is divisible by $\frac{10^n-1}{9}$: the quotient is $\frac{10^n+1}{11}$ and $10^n \equiv (-1)^n = -1 \pmod{11}$. If $n>1$, then $\frac{10^n-1}{9} > 1$ and $\frac{10^n+1}{11} > 1$ so $a_n$ is composite; if $n=1$, then $a_n=1$ which is not prime.

Answer 2

The number $1(01)_n$ is $(100^{n+1}-1)/(100-1)$ by thinking of the number as a geometric series. The numerator factors as $(10^{n+1}+1)(10^{n+1}-1)$. Once $n$ is larger than $1$, one of these two factors will have factors outside of $99$. The $n=1$ case is $101$, which is prime.

If this were base-2, your observations stem from this being nearly a product of a Fermat and a Mersenne number, so perhaps this is the base-10 version of those numbers.

Answer 3

Let $N = 10101....01 = \sum\limits_{k=0}^{n-1} 100^k$ where $n$ are the number of zeros.

$N*99 = N(100 - 1) = \sum\limits_{k=0}^{n} 100^k - \sum\limits_{k=0}^n 100^k = 100^{n} - 1 = 10^{2n} - 1 = (10^{n} - 1)(10^{n} + 1)$

$10^{n} - 1 = 999999..... $($n$ $9$s) $

So $11*N = 111111.....1 *(10000000.....1)$. $11$ is prime so either $11|1111111...1$ (which it does if $n$ is even but not if $n$ is odd) or $11|100000 ...1$ (which it does if $n$ is odd but not if $n$ is even... because a number is divisible by $11$ when the even place decimals add up to the odd place decimals. So the ones are in the $0$ and the $n$ spot so....)

So $N = \frac {111111...... *100000...... 01}{11}$ which is which is composite either by:

$N = \frac {111111.....1}{11}*1000000.....01$ (if $n$ is even) or by:

$N = 111111....111*\frac {10000..... 0 1}{11}$ (if $n$ is odd)

So $N$ is composite....

Unless $111....1 = 11$ (i.e. $n=2$ and $N = 101$) or $1000...1 = 11$ and ... well that would mean $n=0$ and $N = 1$.

....

In retrospect:

$N= 101010101..... 101$ ($n$ ones).

$11N = 11*1010101010......101 = 111111111......111$ ($2n$ ones).

$= 11111111....11*1000000.....1$ ($n$ ones and $n-1$ zeros).

So $N = \frac {11111111....11*1000000.....1}{11}$ which is an integer and as the denominator is prime, the denominator must divide one or the other of the two numerator terms.

So $N$ is composite unless one of the numerator terms is $11$. $1000...01$ can't be $11$ unless $n = 1$ and $N = 1$. $11111....1 = 11$ only if $n=2$ and $N = 101$.

Answer 4

The number is of the final form $(10^{2n} -1)/99$. Where $n \geq 2$.

Lets assume that to be a prime $p$, $p\geq 1$. Then we have $99p = (10^{2n} -1) = (10^n-1)\times (10^n+1)$ with $n\geq 2$.

Now we observe that in R.H.S. we have 2 numbers separated by 2. $(10^n+1)-(10^n-1) = 2$

Hence, we can factorize L.H.S. = $99p = 3^2\times 11\times p$ into 2 factors $a\times b$ where $(a-b)=\pm 2$. All possible pairs (without the obvious symmetric pairs) are listed as follows. $(99,p),(11,9p),(9,11p)$ and $(33,3p),(99p,1)(33p,3)$.

$33p-3 =\pm2 $ and $(33-3p) =\pm2$ can be discarded as $ 3\not| 2$.
$(99p-1)=2$ gives $99p =3$ not possible for integer $p$.

$99-p = \pm2$ gives $p=101$ or 97. $p=101$ $\implies$ n=2 which gives the prime 101.

$p=97$ which is clearly not possible as the number is '1 $mod(100)$' (by definition).

Now we are left with other 2 choices. $(11-9p)= \pm2$ means $9p= 13 $ or $9$. So only possibility is the trivial $p=1$. We discard $9p =13$ as not possible.

Similarly,$(9-11p) =\pm2$ gives $11p = 11$ or $7$. $7$ can be discarded as $11 \not| $ 7. For $11p=11$ we get trivial answer p=1.

Finally, 101 is the only prime of this form.