A simple algorithm to prime factorization












0












$begingroup$


Here a proff:



We know that:



$n!=prod_{P_{i} leq n}p_{i}^{ alpha_{i}(n)}$; where:



$alpha_{i}(n)=sum_{t=1}^{r}[frac{n}{p_{i}^{t}}]$ and $p^{r} leq n < p^{r+1}$.



Then:



$n=frac {n!}{(n-1)!}=prod_{P_{i} leq n}p_{i}^{ beta_{i}(n)}$ (Eq. 1)



Where:



$beta_{i}(n)= alpha_{i}(n)-alpha_{i}(n-1)$



In other words, this method can be used also to found a relation between the prime factirization of n+1 from n, lets see:



$n+1=prod_{P_{i} leq n+1}p_{i}^{ beta_{i}(n+1)}$; where:



$beta_{i}(n+1)= alpha_{i}(n+1)-alpha_{i}(n)$.



Finally, this is the relation:



$beta_{i}(n+1)= alpha_{i}(n+1)-alpha_{i}(n-1)-beta_{i}(n)$.



Example of prime factorization whit this method:



$n=60$



$beta_{i}(60)=sum_{t}^{r} {[frac {60}{p_{i}^{t}}]-[frac {59}{p_{i}^{t}}]}$



Then:



$beta_{1}(60)=sum_{t}^{r} {[frac {60}{2^{t}}]-[frac {59}{2^{t}}]}=2$



$beta_{2}(60)=sum_{t}^{r} {[frac {60}{3^{t}}]-[frac {59}{3^{t}}]}=1$



$beta_{1}(60)=sum_{t}^{r} {[frac {60}{5^{t}}]-[frac {59}{5^{t}}]}=1$



Finally:



$60=2^{2}3^{1}5^{1}$






number-theory







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    What is your question? And what is the point of your proposed algorithm? It involves much more work to calculate what you call $alpha_i(n)$ than it does to calculate $beta_i(n)$ via a naive method (trial division).
    $endgroup$
    – Rob Arthan
    Dec 4 '18 at 23:41












  • $begingroup$
    If you expect that for large numbers , there is a relation between the factorizations of $N$ and $N+1$ that allows quick factorization, if we know the factorizaion of one of the numbers, I must disappoint you.
    $endgroup$
    – Peter
    Dec 5 '18 at 13:05
















0












$begingroup$


Here a proff:



We know that:



$n!=prod_{P_{i} leq n}p_{i}^{ alpha_{i}(n)}$; where:



$alpha_{i}(n)=sum_{t=1}^{r}[frac{n}{p_{i}^{t}}]$ and $p^{r} leq n < p^{r+1}$.



Then:



$n=frac {n!}{(n-1)!}=prod_{P_{i} leq n}p_{i}^{ beta_{i}(n)}$ (Eq. 1)



Where:



$beta_{i}(n)= alpha_{i}(n)-alpha_{i}(n-1)$



In other words, this method can be used also to found a relation between the prime factirization of n+1 from n, lets see:



$n+1=prod_{P_{i} leq n+1}p_{i}^{ beta_{i}(n+1)}$; where:



$beta_{i}(n+1)= alpha_{i}(n+1)-alpha_{i}(n)$.



Finally, this is the relation:



$beta_{i}(n+1)= alpha_{i}(n+1)-alpha_{i}(n-1)-beta_{i}(n)$.



Example of prime factorization whit this method:



$n=60$



$beta_{i}(60)=sum_{t}^{r} {[frac {60}{p_{i}^{t}}]-[frac {59}{p_{i}^{t}}]}$



Then:



$beta_{1}(60)=sum_{t}^{r} {[frac {60}{2^{t}}]-[frac {59}{2^{t}}]}=2$



$beta_{2}(60)=sum_{t}^{r} {[frac {60}{3^{t}}]-[frac {59}{3^{t}}]}=1$



$beta_{1}(60)=sum_{t}^{r} {[frac {60}{5^{t}}]-[frac {59}{5^{t}}]}=1$



Finally:



$60=2^{2}3^{1}5^{1}$






number-theory







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    What is your question? And what is the point of your proposed algorithm? It involves much more work to calculate what you call $alpha_i(n)$ than it does to calculate $beta_i(n)$ via a naive method (trial division).
    $endgroup$
    – Rob Arthan
    Dec 4 '18 at 23:41












  • $begingroup$
    If you expect that for large numbers , there is a relation between the factorizations of $N$ and $N+1$ that allows quick factorization, if we know the factorizaion of one of the numbers, I must disappoint you.
    $endgroup$
    – Peter
    Dec 5 '18 at 13:05














0












0








0





$begingroup$


Here a proff:



We know that:



$n!=prod_{P_{i} leq n}p_{i}^{ alpha_{i}(n)}$; where:



$alpha_{i}(n)=sum_{t=1}^{r}[frac{n}{p_{i}^{t}}]$ and $p^{r} leq n < p^{r+1}$.



Then:



$n=frac {n!}{(n-1)!}=prod_{P_{i} leq n}p_{i}^{ beta_{i}(n)}$ (Eq. 1)



Where:



$beta_{i}(n)= alpha_{i}(n)-alpha_{i}(n-1)$



In other words, this method can be used also to found a relation between the prime factirization of n+1 from n, lets see:



$n+1=prod_{P_{i} leq n+1}p_{i}^{ beta_{i}(n+1)}$; where:



$beta_{i}(n+1)= alpha_{i}(n+1)-alpha_{i}(n)$.



Finally, this is the relation:



$beta_{i}(n+1)= alpha_{i}(n+1)-alpha_{i}(n-1)-beta_{i}(n)$.



Example of prime factorization whit this method:



$n=60$



$beta_{i}(60)=sum_{t}^{r} {[frac {60}{p_{i}^{t}}]-[frac {59}{p_{i}^{t}}]}$



Then:



$beta_{1}(60)=sum_{t}^{r} {[frac {60}{2^{t}}]-[frac {59}{2^{t}}]}=2$



$beta_{2}(60)=sum_{t}^{r} {[frac {60}{3^{t}}]-[frac {59}{3^{t}}]}=1$



$beta_{1}(60)=sum_{t}^{r} {[frac {60}{5^{t}}]-[frac {59}{5^{t}}]}=1$



Finally:



$60=2^{2}3^{1}5^{1}$






number-theory







share|cite|improve this question









$endgroup$




Here a proff:



We know that:



$n!=prod_{P_{i} leq n}p_{i}^{ alpha_{i}(n)}$; where:



$alpha_{i}(n)=sum_{t=1}^{r}[frac{n}{p_{i}^{t}}]$ and $p^{r} leq n < p^{r+1}$.



Then:



$n=frac {n!}{(n-1)!}=prod_{P_{i} leq n}p_{i}^{ beta_{i}(n)}$ (Eq. 1)



Where:



$beta_{i}(n)= alpha_{i}(n)-alpha_{i}(n-1)$



In other words, this method can be used also to found a relation between the prime factirization of n+1 from n, lets see:



$n+1=prod_{P_{i} leq n+1}p_{i}^{ beta_{i}(n+1)}$; where:



$beta_{i}(n+1)= alpha_{i}(n+1)-alpha_{i}(n)$.



Finally, this is the relation:



$beta_{i}(n+1)= alpha_{i}(n+1)-alpha_{i}(n-1)-beta_{i}(n)$.



Example of prime factorization whit this method:



$n=60$



$beta_{i}(60)=sum_{t}^{r} {[frac {60}{p_{i}^{t}}]-[frac {59}{p_{i}^{t}}]}$



Then:



$beta_{1}(60)=sum_{t}^{r} {[frac {60}{2^{t}}]-[frac {59}{2^{t}}]}=2$



$beta_{2}(60)=sum_{t}^{r} {[frac {60}{3^{t}}]-[frac {59}{3^{t}}]}=1$



$beta_{1}(60)=sum_{t}^{r} {[frac {60}{5^{t}}]-[frac {59}{5^{t}}]}=1$



Finally:



$60=2^{2}3^{1}5^{1}$






number-theory




number-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 4 '18 at 23:32









Mauricio AreizaMauricio Areiza

143




143








  • 4




    $begingroup$
    What is your question? And what is the point of your proposed algorithm? It involves much more work to calculate what you call $alpha_i(n)$ than it does to calculate $beta_i(n)$ via a naive method (trial division).
    $endgroup$
    – Rob Arthan
    Dec 4 '18 at 23:41












  • $begingroup$
    If you expect that for large numbers , there is a relation between the factorizations of $N$ and $N+1$ that allows quick factorization, if we know the factorizaion of one of the numbers, I must disappoint you.
    $endgroup$
    – Peter
    Dec 5 '18 at 13:05














  • 4




    $begingroup$
    What is your question? And what is the point of your proposed algorithm? It involves much more work to calculate what you call $alpha_i(n)$ than it does to calculate $beta_i(n)$ via a naive method (trial division).
    $endgroup$
    – Rob Arthan
    Dec 4 '18 at 23:41












  • $begingroup$
    If you expect that for large numbers , there is a relation between the factorizations of $N$ and $N+1$ that allows quick factorization, if we know the factorizaion of one of the numbers, I must disappoint you.
    $endgroup$
    – Peter
    Dec 5 '18 at 13:05








4




4




$begingroup$
What is your question? And what is the point of your proposed algorithm? It involves much more work to calculate what you call $alpha_i(n)$ than it does to calculate $beta_i(n)$ via a naive method (trial division).
$endgroup$
– Rob Arthan
Dec 4 '18 at 23:41






$begingroup$
What is your question? And what is the point of your proposed algorithm? It involves much more work to calculate what you call $alpha_i(n)$ than it does to calculate $beta_i(n)$ via a naive method (trial division).
$endgroup$
– Rob Arthan
Dec 4 '18 at 23:41














$begingroup$
If you expect that for large numbers , there is a relation between the factorizations of $N$ and $N+1$ that allows quick factorization, if we know the factorizaion of one of the numbers, I must disappoint you.
$endgroup$
– Peter
Dec 5 '18 at 13:05




$begingroup$
If you expect that for large numbers , there is a relation between the factorizations of $N$ and $N+1$ that allows quick factorization, if we know the factorizaion of one of the numbers, I must disappoint you.
$endgroup$
– Peter
Dec 5 '18 at 13:05










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