Primes $p$ of the form $p=frac7^q+17^q-n^2+1$, I ask some questionsPrimes of the form $frac7^q+17^m+1$, where q is a positive integer and $m=q-n^2$Is it possible for a number in form $1987^k-1$ to end with 1987 zeros? Also few questions about number theory in general.Some questions about Goldbach's conjectureWhat numbers can be expressed with the following expression?Generating pairs of primes from the 2 previous primes.Find all primes of the form $3^n -1$Proving there are infinite primes of some form $a+nd$Finding primes by summing the first n consecutive primesPrimes in intervals of special form and search for the minimal $k$ (if it exists)Can you propose a conjectural $textUpper bound(x)$ for the counting function of a sequence of primes arising from the Eratosthenes sieve?The hunting of “missing primes”
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Primes $p$ of the form $p=frac7^q+17^q-n^2+1$, I ask some questions
Primes of the form $frac7^q+17^m+1$, where q is a positive integer and $m=q-n^2$Is it possible for a number in form $1987^k-1$ to end with 1987 zeros? Also few questions about number theory in general.Some questions about Goldbach's conjectureWhat numbers can be expressed with the following expression?Generating pairs of primes from the 2 previous primes.Find all primes of the form $3^n -1$Proving there are infinite primes of some form $a+nd$Finding primes by summing the first n consecutive primesPrimes in intervals of special form and search for the minimal $k$ (if it exists)Can you propose a conjectural $textUpper bound(x)$ for the counting function of a sequence of primes arising from the Eratosthenes sieve?The hunting of “missing primes”
$begingroup$
In a previous post I asked for primes $p$ of the form: $$p=frac7^q+17^q-n^2+1,$$ where $q$ and $n$ are positive integers.
The solutions found up to $q=5000$ are $[1,17,4]$, $[8,24,4]$, $[2,38,6]$ and $[4,148,12]$. In brackets the second number is $q$, the first number is $q-n^2$ and the third number is $n$.
Do you believe that the next solution is extremely huge or even far beyond the programs and calculators capacities?
Do you believe that the number of these primes is not infinite?
Do you believe that a prime of this type exists such that $q-n^2neq 2^k$ where $k$ is an integer $geqslant 0$?
Here is the link to the previous question
number-theory prime-numbers
$endgroup$
add a comment |
$begingroup$
In a previous post I asked for primes $p$ of the form: $$p=frac7^q+17^q-n^2+1,$$ where $q$ and $n$ are positive integers.
The solutions found up to $q=5000$ are $[1,17,4]$, $[8,24,4]$, $[2,38,6]$ and $[4,148,12]$. In brackets the second number is $q$, the first number is $q-n^2$ and the third number is $n$.
Do you believe that the next solution is extremely huge or even far beyond the programs and calculators capacities?
Do you believe that the number of these primes is not infinite?
Do you believe that a prime of this type exists such that $q-n^2neq 2^k$ where $k$ is an integer $geqslant 0$?
Here is the link to the previous question
number-theory prime-numbers
$endgroup$
$begingroup$
Why are there 3 numbers in the square brackets?
$endgroup$
– Pink Panther
Mar 21 at 8:05
$begingroup$
@Pink Panter the second number is q, the first number is $q-n^2$ and the third number is $n$.
$endgroup$
– homunculus
Mar 21 at 8:07
add a comment |
$begingroup$
In a previous post I asked for primes $p$ of the form: $$p=frac7^q+17^q-n^2+1,$$ where $q$ and $n$ are positive integers.
The solutions found up to $q=5000$ are $[1,17,4]$, $[8,24,4]$, $[2,38,6]$ and $[4,148,12]$. In brackets the second number is $q$, the first number is $q-n^2$ and the third number is $n$.
Do you believe that the next solution is extremely huge or even far beyond the programs and calculators capacities?
Do you believe that the number of these primes is not infinite?
Do you believe that a prime of this type exists such that $q-n^2neq 2^k$ where $k$ is an integer $geqslant 0$?
Here is the link to the previous question
number-theory prime-numbers
$endgroup$
In a previous post I asked for primes $p$ of the form: $$p=frac7^q+17^q-n^2+1,$$ where $q$ and $n$ are positive integers.
The solutions found up to $q=5000$ are $[1,17,4]$, $[8,24,4]$, $[2,38,6]$ and $[4,148,12]$. In brackets the second number is $q$, the first number is $q-n^2$ and the third number is $n$.
Do you believe that the next solution is extremely huge or even far beyond the programs and calculators capacities?
Do you believe that the number of these primes is not infinite?
Do you believe that a prime of this type exists such that $q-n^2neq 2^k$ where $k$ is an integer $geqslant 0$?
Here is the link to the previous question
number-theory prime-numbers
number-theory prime-numbers
edited Mar 21 at 8:50
Andrews
1,2812422
1,2812422
asked Mar 21 at 8:00
homunculushomunculus
1869
1869
$begingroup$
Why are there 3 numbers in the square brackets?
$endgroup$
– Pink Panther
Mar 21 at 8:05
$begingroup$
@Pink Panter the second number is q, the first number is $q-n^2$ and the third number is $n$.
$endgroup$
– homunculus
Mar 21 at 8:07
add a comment |
$begingroup$
Why are there 3 numbers in the square brackets?
$endgroup$
– Pink Panther
Mar 21 at 8:05
$begingroup$
@Pink Panter the second number is q, the first number is $q-n^2$ and the third number is $n$.
$endgroup$
– homunculus
Mar 21 at 8:07
$begingroup$
Why are there 3 numbers in the square brackets?
$endgroup$
– Pink Panther
Mar 21 at 8:05
$begingroup$
Why are there 3 numbers in the square brackets?
$endgroup$
– Pink Panther
Mar 21 at 8:05
$begingroup$
@Pink Panter the second number is q, the first number is $q-n^2$ and the third number is $n$.
$endgroup$
– homunculus
Mar 21 at 8:07
$begingroup$
@Pink Panter the second number is q, the first number is $q-n^2$ and the third number is $n$.
$endgroup$
– homunculus
Mar 21 at 8:07
add a comment |
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$begingroup$
Why are there 3 numbers in the square brackets?
$endgroup$
– Pink Panther
Mar 21 at 8:05
$begingroup$
@Pink Panter the second number is q, the first number is $q-n^2$ and the third number is $n$.
$endgroup$
– homunculus
Mar 21 at 8:07