Any good way to calculate $frac {alpha ^ n - 1 } {alpha - 1} pmod{c}$
4
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
edited yesterday
rtybase
10.4k21433
asked 2 days ago
satvik choudhary
215
New contributor
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add a comment |
4
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
edited yesterday
rtybase
10.4k21433
asked 2 days ago
satvik choudhary
215
New contributor
satvik choudhary is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
If $c$ divides $alpha-1$ with multiplicity $k$, you can compute $alpha^n-1pmod{p^{k+1}}$ and then divide by $alpha-1$.
– Wojowu
2 days ago
add a comment |
4
4
4
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
edited yesterday
rtybase
10.4k21433
asked 2 days ago
satvik choudhary
215
New contributor
satvik choudhary is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
edited yesterday
rtybase
10.4k21433
asked 2 days ago
satvik choudhary
215
New contributor
satvik choudhary is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
rtybase
10.4k21433
asked 2 days ago
satvik choudhary
215
New contributor
satvik choudhary is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
rtybase
10.4k21433
edited yesterday
rtybase
10.4k21433
edited yesterday
rtybase
10.4k21433
10.4k21433
asked 2 days ago
satvik choudhary
215
New contributor
satvik choudhary is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
satvik choudhary
215
asked 2 days ago
satvik choudhary
215
215
New contributor
satvik choudhary is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
satvik choudhary is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
satvik choudhary is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
If $c$ divides $alpha-1$ with multiplicity $k$, you can compute $alpha^n-1pmod{p^{k+1}}$ and then divide by $alpha-1$.
– Wojowu
2 days ago
add a comment |
1
If $c$ divides $alpha-1$ with multiplicity $k$, you can compute $alpha^n-1pmod{p^{k+1}}$ and then divide by $alpha-1$.
– Wojowu
2 days ago
1
1
If $c$ divides $alpha-1$ with multiplicity $k$, you can compute $alpha^n-1pmod{p^{k+1}}$ and then divide by $alpha-1$.
– Wojowu
2 days ago
If $c$ divides $alpha-1$ with multiplicity $k$, you can compute $alpha^n-1pmod{p^{k+1}}$ and then divide by $alpha-1$.
– Wojowu
2 days ago
add a comment |
1 Answer
1
active
oldest
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2
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.
answered 2 days ago
W-t-P
92359
Its too slow to be just linearly calculated with n ~ 1e9
– satvik choudhary
yesterday
A better way would be to go like $ S_k := S_{frac {k}{2}} + alpha ^ {k / 2} S_{k - frac{k}{2}} $
– satvik choudhary
yesterday
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.)
– W-t-P
20 hours ago
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
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2
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.
answered 2 days ago
W-t-P
92359
Its too slow to be just linearly calculated with n ~ 1e9
– satvik choudhary
yesterday
A better way would be to go like $ S_k := S_{frac {k}{2}} + alpha ^ {k / 2} S_{k - frac{k}{2}} $
– satvik choudhary
yesterday
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.)
– W-t-P
20 hours ago
add a comment |
2
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.
answered 2 days ago
W-t-P
92359
Its too slow to be just linearly calculated with n ~ 1e9
– satvik choudhary
yesterday
A better way would be to go like $ S_k := S_{frac {k}{2}} + alpha ^ {k / 2} S_{k - frac{k}{2}} $
– satvik choudhary
yesterday
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.)
– W-t-P
20 hours ago
add a comment |
2
2
2
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.
answered 2 days ago
W-t-P
92359
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.
answered 2 days ago
W-t-P
92359
answered 2 days ago
W-t-P
92359
answered 2 days ago
W-t-P
92359
answered 2 days ago
W-t-P
92359
92359
Its too slow to be just linearly calculated with n ~ 1e9
– satvik choudhary
yesterday
A better way would be to go like $ S_k := S_{frac {k}{2}} + alpha ^ {k / 2} S_{k - frac{k}{2}} $
– satvik choudhary
yesterday
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.)
– W-t-P
20 hours ago
add a comment |
Its too slow to be just linearly calculated with n ~ 1e9
– satvik choudhary
yesterday
A better way would be to go like $ S_k := S_{frac {k}{2}} + alpha ^ {k / 2} S_{k - frac{k}{2}} $
– satvik choudhary
yesterday
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.)
– W-t-P
20 hours ago
Its too slow to be just linearly calculated with n ~ 1e9
– satvik choudhary
yesterday
Its too slow to be just linearly calculated with n ~ 1e9
– satvik choudhary
yesterday
A better way would be to go like $ S_k := S_{frac {k}{2}} + alpha ^ {k / 2} S_{k - frac{k}{2}} $
– satvik choudhary
yesterday
A better way would be to go like $ S_k := S_{frac {k}{2}} + alpha ^ {k / 2} S_{k - frac{k}{2}} $
– satvik choudhary
yesterday
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.)
– W-t-P
20 hours ago
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.)
– W-t-P
20 hours ago
add a comment |
satvik choudhary is a new contributor. Be nice, and check out our Code of Conduct.
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1
If $c$ divides $alpha-1$ with multiplicity $k$, you can compute $alpha^n-1pmod{p^{k+1}}$ and then divide by $alpha-1$.
– Wojowu
2 days ago