Any good way to calculate $frac {alpha ^ n - 1 } {alpha - 1} pmod{c}$












4












$begingroup$


I tried by multiplying modular inverse of denominator to the numerator and then taking modulo $c$, but there are problems when the inverse does not exist.



So is there a good way to solve this problem.



Constraints
$$ 1 le alpha le 1e9 $$
$c$ is a prime
$$ 1 le n le 1e9 $$






number-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If $c$ divides $alpha-1$ with multiplicity $k$, you can compute $alpha^n-1pmod{p^{k+1}}$ and then divide by $alpha-1$.
    $endgroup$
    – Wojowu
    Jan 3 at 17:04
















4












$begingroup$


I tried by multiplying modular inverse of denominator to the numerator and then taking modulo $c$, but there are problems when the inverse does not exist.



So is there a good way to solve this problem.



Constraints
$$ 1 le alpha le 1e9 $$
$c$ is a prime
$$ 1 le n le 1e9 $$






number-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If $c$ divides $alpha-1$ with multiplicity $k$, you can compute $alpha^n-1pmod{p^{k+1}}$ and then divide by $alpha-1$.
    $endgroup$
    – Wojowu
    Jan 3 at 17:04














4












4








4





$begingroup$


I tried by multiplying modular inverse of denominator to the numerator and then taking modulo $c$, but there are problems when the inverse does not exist.



So is there a good way to solve this problem.



Constraints
$$ 1 le alpha le 1e9 $$
$c$ is a prime
$$ 1 le n le 1e9 $$






number-theory







share|cite|improve this question











$endgroup$




I tried by multiplying modular inverse of denominator to the numerator and then taking modulo $c$, but there are problems when the inverse does not exist.



So is there a good way to solve this problem.



Constraints
$$ 1 le alpha le 1e9 $$
$c$ is a prime
$$ 1 le n le 1e9 $$






number-theory




number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 21:45









rtybase

11.2k21533




11.2k21533










asked Jan 3 at 16:59









satvik choudharysatvik choudhary

215




215








  • 1




    $begingroup$
    If $c$ divides $alpha-1$ with multiplicity $k$, you can compute $alpha^n-1pmod{p^{k+1}}$ and then divide by $alpha-1$.
    $endgroup$
    – Wojowu
    Jan 3 at 17:04














  • 1




    $begingroup$
    If $c$ divides $alpha-1$ with multiplicity $k$, you can compute $alpha^n-1pmod{p^{k+1}}$ and then divide by $alpha-1$.
    $endgroup$
    – Wojowu
    Jan 3 at 17:04








1




1




$begingroup$
If $c$ divides $alpha-1$ with multiplicity $k$, you can compute $alpha^n-1pmod{p^{k+1}}$ and then divide by $alpha-1$.
$endgroup$
– Wojowu
Jan 3 at 17:04




$begingroup$
If $c$ divides $alpha-1$ with multiplicity $k$, you can compute $alpha^n-1pmod{p^{k+1}}$ and then divide by $alpha-1$.
$endgroup$
– Wojowu
Jan 3 at 17:04










1 Answer
1






active

oldest

votes


















2












$begingroup$

Set $S_0:=1$ and then recursively $S_k:=alpha S_{k-1}+1 pmod c$ for all $k=1,dotsc,n-1$. The last value $S_{n-1}$ is what you seek.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Its too slow to be just linearly calculated with n ~ 1e9
    $endgroup$
    – satvik choudhary
    Jan 4 at 10:19










  • $begingroup$
    A better way would be to go like $ S_k := S_{frac {k}{2}} + alpha ^ {k / 2} S_{k - frac{k}{2}} $
    $endgroup$
    – satvik choudhary
    Jan 4 at 10:29










  • $begingroup$
    Sure, as long as running time is an issue. (From your question I understood that you are mostly struggling with the situation where the denominator is not invertible.)
    $endgroup$
    – W-t-P
    Jan 5 at 9:01











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









2












$begingroup$

Set $S_0:=1$ and then recursively $S_k:=alpha S_{k-1}+1 pmod c$ for all $k=1,dotsc,n-1$. The last value $S_{n-1}$ is what you seek.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Its too slow to be just linearly calculated with n ~ 1e9
    $endgroup$
    – satvik choudhary
    Jan 4 at 10:19










  • $begingroup$
    A better way would be to go like $ S_k := S_{frac {k}{2}} + alpha ^ {k / 2} S_{k - frac{k}{2}} $
    $endgroup$
    – satvik choudhary
    Jan 4 at 10:29










  • $begingroup$
    Sure, as long as running time is an issue. (From your question I understood that you are mostly struggling with the situation where the denominator is not invertible.)
    $endgroup$
    – W-t-P
    Jan 5 at 9:01
















2












$begingroup$

Set $S_0:=1$ and then recursively $S_k:=alpha S_{k-1}+1 pmod c$ for all $k=1,dotsc,n-1$. The last value $S_{n-1}$ is what you seek.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Its too slow to be just linearly calculated with n ~ 1e9
    $endgroup$
    – satvik choudhary
    Jan 4 at 10:19










  • $begingroup$
    A better way would be to go like $ S_k := S_{frac {k}{2}} + alpha ^ {k / 2} S_{k - frac{k}{2}} $
    $endgroup$
    – satvik choudhary
    Jan 4 at 10:29










  • $begingroup$
    Sure, as long as running time is an issue. (From your question I understood that you are mostly struggling with the situation where the denominator is not invertible.)
    $endgroup$
    – W-t-P
    Jan 5 at 9:01














2












2








2





$begingroup$

Set $S_0:=1$ and then recursively $S_k:=alpha S_{k-1}+1 pmod c$ for all $k=1,dotsc,n-1$. The last value $S_{n-1}$ is what you seek.






share|cite|improve this answer









$endgroup$



Set $S_0:=1$ and then recursively $S_k:=alpha S_{k-1}+1 pmod c$ for all $k=1,dotsc,n-1$. The last value $S_{n-1}$ is what you seek.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 3 at 18:21









W-t-PW-t-P

1,302611




1,302611












  • $begingroup$
    Its too slow to be just linearly calculated with n ~ 1e9
    $endgroup$
    – satvik choudhary
    Jan 4 at 10:19










  • $begingroup$
    A better way would be to go like $ S_k := S_{frac {k}{2}} + alpha ^ {k / 2} S_{k - frac{k}{2}} $
    $endgroup$
    – satvik choudhary
    Jan 4 at 10:29










  • $begingroup$
    Sure, as long as running time is an issue. (From your question I understood that you are mostly struggling with the situation where the denominator is not invertible.)
    $endgroup$
    – W-t-P
    Jan 5 at 9:01


















  • $begingroup$
    Its too slow to be just linearly calculated with n ~ 1e9
    $endgroup$
    – satvik choudhary
    Jan 4 at 10:19










  • $begingroup$
    A better way would be to go like $ S_k := S_{frac {k}{2}} + alpha ^ {k / 2} S_{k - frac{k}{2}} $
    $endgroup$
    – satvik choudhary
    Jan 4 at 10:29










  • $begingroup$
    Sure, as long as running time is an issue. (From your question I understood that you are mostly struggling with the situation where the denominator is not invertible.)
    $endgroup$
    – W-t-P
    Jan 5 at 9:01
















$begingroup$
Its too slow to be just linearly calculated with n ~ 1e9
$endgroup$
– satvik choudhary
Jan 4 at 10:19




$begingroup$
Its too slow to be just linearly calculated with n ~ 1e9
$endgroup$
– satvik choudhary
Jan 4 at 10:19












$begingroup$
A better way would be to go like $ S_k := S_{frac {k}{2}} + alpha ^ {k / 2} S_{k - frac{k}{2}} $
$endgroup$
– satvik choudhary
Jan 4 at 10:29




$begingroup$
A better way would be to go like $ S_k := S_{frac {k}{2}} + alpha ^ {k / 2} S_{k - frac{k}{2}} $
$endgroup$
– satvik choudhary
Jan 4 at 10:29












$begingroup$
Sure, as long as running time is an issue. (From your question I understood that you are mostly struggling with the situation where the denominator is not invertible.)
$endgroup$
– W-t-P
Jan 5 at 9:01




$begingroup$
Sure, as long as running time is an issue. (From your question I understood that you are mostly struggling with the situation where the denominator is not invertible.)
$endgroup$
– W-t-P
Jan 5 at 9:01


















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