Bounds on primes with Gilbreath's Conjecture
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Gilbreath's conjecture is a famous conjecture on prime numbers that states that when we take the prime number sequence and perform iterative subtraction, we repeatedly get the number $1$. A link to Wikipedia is provided below:
https://en.wikipedia.org/wiki/Gilbreath%27s_conjecture
Is it possible to get any bound on primes conditionally on this conjecture, that is strong? It seems that the bounds look alike the statement of Bertrand's Postulate, if we substitute the values of powers of $2$, instead of primes, where the statement of the conjecture holds trivially. Thus, it might be likely that some good bound can be obtained.
This bound looks quite strong as when we add the coefficients of the primes, we get $0$. What is the bound we obtain on the $n$th prime? Can this bound be converted to closed form? And if so, does the bound actually hold? (i.e. if the bound is $p_{n+1} < 2p_n$, the bound holds unconditionally.)
number-theory prime-numbers
asked Dec 30 '18 at 16:36
HaranHaran
1,332524
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add a comment |
$begingroup$
Gilbreath's conjecture is a famous conjecture on prime numbers that states that when we take the prime number sequence and perform iterative subtraction, we repeatedly get the number $1$. A link to Wikipedia is provided below:
https://en.wikipedia.org/wiki/Gilbreath%27s_conjecture
Is it possible to get any bound on primes conditionally on this conjecture, that is strong? It seems that the bounds look alike the statement of Bertrand's Postulate, if we substitute the values of powers of $2$, instead of primes, where the statement of the conjecture holds trivially. Thus, it might be likely that some good bound can be obtained.
This bound looks quite strong as when we add the coefficients of the primes, we get $0$. What is the bound we obtain on the $n$th prime? Can this bound be converted to closed form? And if so, does the bound actually hold? (i.e. if the bound is $p_{n+1} < 2p_n$, the bound holds unconditionally.)
number-theory prime-numbers
asked Dec 30 '18 at 16:36
HaranHaran
1,332524
$endgroup$
$begingroup$
You can get 1 with almost everything. The sequence of odd numbers begining with 2 (2,3,5,7,9,11,....) or the sequence (2,3,5,9,17,...$a_{n+1}=2cdot a_n-1$) also gives 1. Like Richard said, there is nothing special about primes in this conjecture. Just that "the primes grow slowly and are reasonably distributed". But indeed, perhaps it can give you some bound on primes althought I have doubts ($a_{n+1}=2cdot a_n-1$ is already Bertrand's postulate).
$endgroup$
– Collag3n
Dec 30 '18 at 18:32
add a comment |
$begingroup$
Gilbreath's conjecture is a famous conjecture on prime numbers that states that when we take the prime number sequence and perform iterative subtraction, we repeatedly get the number $1$. A link to Wikipedia is provided below:
https://en.wikipedia.org/wiki/Gilbreath%27s_conjecture
Is it possible to get any bound on primes conditionally on this conjecture, that is strong? It seems that the bounds look alike the statement of Bertrand's Postulate, if we substitute the values of powers of $2$, instead of primes, where the statement of the conjecture holds trivially. Thus, it might be likely that some good bound can be obtained.
This bound looks quite strong as when we add the coefficients of the primes, we get $0$. What is the bound we obtain on the $n$th prime? Can this bound be converted to closed form? And if so, does the bound actually hold? (i.e. if the bound is $p_{n+1} < 2p_n$, the bound holds unconditionally.)
number-theory prime-numbers
asked Dec 30 '18 at 16:36
HaranHaran
1,332524
$endgroup$
Gilbreath's conjecture is a famous conjecture on prime numbers that states that when we take the prime number sequence and perform iterative subtraction, we repeatedly get the number $1$. A link to Wikipedia is provided below:
https://en.wikipedia.org/wiki/Gilbreath%27s_conjecture
Is it possible to get any bound on primes conditionally on this conjecture, that is strong? It seems that the bounds look alike the statement of Bertrand's Postulate, if we substitute the values of powers of $2$, instead of primes, where the statement of the conjecture holds trivially. Thus, it might be likely that some good bound can be obtained.
This bound looks quite strong as when we add the coefficients of the primes, we get $0$. What is the bound we obtain on the $n$th prime? Can this bound be converted to closed form? And if so, does the bound actually hold? (i.e. if the bound is $p_{n+1} < 2p_n$, the bound holds unconditionally.)
number-theory prime-numbers
number-theory prime-numbers
asked Dec 30 '18 at 16:36
HaranHaran
1,332524
asked Dec 30 '18 at 16:36
HaranHaran
1,332524
asked Dec 30 '18 at 16:36
HaranHaran
1,332524
asked Dec 30 '18 at 16:36
HaranHaran
1,332524
asked Dec 30 '18 at 16:36
HaranHaran
1,332524
1,332524
$begingroup$
You can get 1 with almost everything. The sequence of odd numbers begining with 2 (2,3,5,7,9,11,....) or the sequence (2,3,5,9,17,...$a_{n+1}=2cdot a_n-1$) also gives 1. Like Richard said, there is nothing special about primes in this conjecture. Just that "the primes grow slowly and are reasonably distributed". But indeed, perhaps it can give you some bound on primes althought I have doubts ($a_{n+1}=2cdot a_n-1$ is already Bertrand's postulate).
$endgroup$
– Collag3n
Dec 30 '18 at 18:32
add a comment |
$begingroup$
You can get 1 with almost everything. The sequence of odd numbers begining with 2 (2,3,5,7,9,11,....) or the sequence (2,3,5,9,17,...$a_{n+1}=2cdot a_n-1$) also gives 1. Like Richard said, there is nothing special about primes in this conjecture. Just that "the primes grow slowly and are reasonably distributed". But indeed, perhaps it can give you some bound on primes althought I have doubts ($a_{n+1}=2cdot a_n-1$ is already Bertrand's postulate).
$endgroup$
– Collag3n
Dec 30 '18 at 18:32
$begingroup$
You can get 1 with almost everything. The sequence of odd numbers begining with 2 (2,3,5,7,9,11,....) or the sequence (2,3,5,9,17,...$a_{n+1}=2cdot a_n-1$) also gives 1. Like Richard said, there is nothing special about primes in this conjecture. Just that "the primes grow slowly and are reasonably distributed". But indeed, perhaps it can give you some bound on primes althought I have doubts ($a_{n+1}=2cdot a_n-1$ is already Bertrand's postulate).
$endgroup$
– Collag3n
Dec 30 '18 at 18:32
$begingroup$
You can get 1 with almost everything. The sequence of odd numbers begining with 2 (2,3,5,7,9,11,....) or the sequence (2,3,5,9,17,...$a_{n+1}=2cdot a_n-1$) also gives 1. Like Richard said, there is nothing special about primes in this conjecture. Just that "the primes grow slowly and are reasonably distributed". But indeed, perhaps it can give you some bound on primes althought I have doubts ($a_{n+1}=2cdot a_n-1$ is already Bertrand's postulate).
$endgroup$
– Collag3n
Dec 30 '18 at 18:32
add a comment |
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$begingroup$
You can get 1 with almost everything. The sequence of odd numbers begining with 2 (2,3,5,7,9,11,....) or the sequence (2,3,5,9,17,...$a_{n+1}=2cdot a_n-1$) also gives 1. Like Richard said, there is nothing special about primes in this conjecture. Just that "the primes grow slowly and are reasonably distributed". But indeed, perhaps it can give you some bound on primes althought I have doubts ($a_{n+1}=2cdot a_n-1$ is already Bertrand's postulate).
$endgroup$
– Collag3n
Dec 30 '18 at 18:32