Mixing
Finding the remainder of an exhausted $m!$
For any prime $p$, let $n_p(m)$ denote the exponent of $p$ in the factorisation of $m!$, i.e. $m!=p^{n_p(m)}cdot k$ with $pnotmid k$.
I wonder if there is a general formula for $frac{m!}{p^{n_p(m)}}$ modulo $p$?
I could not prove but I believe that the frequencies of $1,2,...,p-1$ are equal.
modular-arithmetic factorial
asked Jul 6 '16 at 21:10
Levent
2,688925
add a comment |
For any prime $p$, let $n_p(m)$ denote the exponent of $p$ in the factorisation of $m!$, i.e. $m!=p^{n_p(m)}cdot k$ with $pnotmid k$.
I wonder if there is a general formula for $frac{m!}{p^{n_p(m)}}$ modulo $p$?
I could not prove but I believe that the frequencies of $1,2,...,p-1$ are equal.
modular-arithmetic factorial
asked Jul 6 '16 at 21:10
Levent
2,688925
"frequencies" in what sense? across all higher factorials?
– Joffan
Jul 6 '16 at 21:50
I simply meant when I take $m=50$ and $p=3,5$ or $7$, the number of occurrences of $1,2,...,p-1$ as remainders are almost equal.
– Levent
Jul 6 '16 at 21:54
You can't talk about a frequency when you only look at one value of $m$.
– Joffan
Jul 6 '16 at 22:42
Sorry, what I mean is when I take $m=2$ up to $50$.
– Levent
Jul 6 '16 at 22:51
add a comment |
For any prime $p$, let $n_p(m)$ denote the exponent of $p$ in the factorisation of $m!$, i.e. $m!=p^{n_p(m)}cdot k$ with $pnotmid k$.
I wonder if there is a general formula for $frac{m!}{p^{n_p(m)}}$ modulo $p$?
I could not prove but I believe that the frequencies of $1,2,...,p-1$ are equal.
modular-arithmetic factorial
asked Jul 6 '16 at 21:10
Levent
2,688925
For any prime $p$, let $n_p(m)$ denote the exponent of $p$ in the factorisation of $m!$, i.e. $m!=p^{n_p(m)}cdot k$ with $pnotmid k$.
I wonder if there is a general formula for $frac{m!}{p^{n_p(m)}}$ modulo $p$?
I could not prove but I believe that the frequencies of $1,2,...,p-1$ are equal.
modular-arithmetic factorial
modular-arithmetic factorial
asked Jul 6 '16 at 21:10
Levent
2,688925
asked Jul 6 '16 at 21:10
Levent
2,688925
edited Nov 19 '18 at 22:44
asked Jul 6 '16 at 21:10
Levent
2,688925
asked Jul 6 '16 at 21:10
Levent
2,688925
asked Jul 6 '16 at 21:10
Levent
2,688925
2,688925
"frequencies" in what sense? across all higher factorials?
– Joffan
Jul 6 '16 at 21:50
I simply meant when I take $m=50$ and $p=3,5$ or $7$, the number of occurrences of $1,2,...,p-1$ as remainders are almost equal.
– Levent
Jul 6 '16 at 21:54
You can't talk about a frequency when you only look at one value of $m$.
– Joffan
Jul 6 '16 at 22:42
Sorry, what I mean is when I take $m=2$ up to $50$.
– Levent
Jul 6 '16 at 22:51
add a comment |
"frequencies" in what sense? across all higher factorials?
– Joffan
Jul 6 '16 at 21:50
I simply meant when I take $m=50$ and $p=3,5$ or $7$, the number of occurrences of $1,2,...,p-1$ as remainders are almost equal.
– Levent
Jul 6 '16 at 21:54
You can't talk about a frequency when you only look at one value of $m$.
– Joffan
Jul 6 '16 at 22:42
Sorry, what I mean is when I take $m=2$ up to $50$.
– Levent
Jul 6 '16 at 22:51
"frequencies" in what sense? across all higher factorials?
– Joffan
Jul 6 '16 at 21:50
"frequencies" in what sense? across all higher factorials?
– Joffan
Jul 6 '16 at 21:50
I simply meant when I take $m=50$ and $p=3,5$ or $7$, the number of occurrences of $1,2,...,p-1$ as remainders are almost equal.
– Levent
Jul 6 '16 at 21:54
I simply meant when I take $m=50$ and $p=3,5$ or $7$, the number of occurrences of $1,2,...,p-1$ as remainders are almost equal.
– Levent
Jul 6 '16 at 21:54
You can't talk about a frequency when you only look at one value of $m$.
– Joffan
Jul 6 '16 at 22:42
You can't talk about a frequency when you only look at one value of $m$.
– Joffan
Jul 6 '16 at 22:42
Sorry, what I mean is when I take $m=2$ up to $50$.
– Levent
Jul 6 '16 at 22:51
Sorry, what I mean is when I take $m=2$ up to $50$.
– Levent
Jul 6 '16 at 22:51
add a comment |
3 Answers
3
active
oldest
votes
Does not seem to be that simple as we let things run high; I did primes 5, 7, 11. Seems reasonable to suggest 1, p-1 are the same and highest, after that not clear, although 5, 7 seem symmetric. I ran 11 pretty high, I guess if a+b = 11 then their counts are similar.
1 count 1330 2 count 1169 3 count 1169 4 count 1332
1 count 2570 2 count 2220 3 count 2230 4 count 2210 5 count 2220 6 count 2550
1 count 121894 2 count 115087 3 count 99710 4 count 100842 5 count 113429
6 count 112909 7 count 99693 8 count 99050 9 count 114123 10 count 123263
int main()
{
int p = 11;
int count[p] ;
int fac = 1;
for(int i = 1; i < p; ++i) count[i] = 0;
for(int i = 1; i <= 100000 * p; ++i)
{
int n = i;
while ( n % p == 0 ) n /= p;
n %= p;
fac *= n;
fac %= p;
count[fac] = 1 + count[fac];
cout << i << " " << fac << " count " << count[fac] << endl;
}
for(int i = 1; i < p; ++i) cout << i << " count " << count[i] << " " ;
cout << endl << endl;
return 0 ;
}
answered Jul 6 '16 at 22:15
Will Jagy
101k599199
1
I had similar results on those primes, but on 13, 17, 19 a result of "1" stayed high but the remaining values started to flatten out (testing to 30000)
– Joffan
Jul 6 '16 at 22:38
add a comment |
With $k$ as in the question, define $F(m)$ to be the residue of $k$ modulo $p$.
A formula
Let $m=a+bp$, where $0le ale p-1$. Then, modulo $p$, $F(m)$ is congruent to $a!(p-1)!^bF(b)$. By Wilson's Theorem this simplifies to $a!(-1)^bF(b)$.
So, in general, let $$m=a_0+a_1p+a_2p^2+...+a_{2n-1}p^{2n-1},0le a_ile p-1.$$
Then, for $A=a_1+a_3+...+a_{2n-1}$, $F(m)$ is congruent, modulo $p$, to
$$a_0!a_1!...a_{2n-1}!(-1)^A.$$
Frequencies
The formula shows that $F(m)$ is the product modulo $p$ $$F(a_0+a_1p) F(a_2+a_3p)...F(a_{2n-2}+a_{2n-1}p).$$
Therefore the frequencies for the $p^2$numbers $0!,1!,...,(p^2-1)!$ determine the frequencies for the $p^4$numbers up to $(p^4-1)!$ etc. etc.
For example, modulo 3 there are 4 1s and 5 2s from 0! to 8!
So up to 80! there are $4^2+5^2=41$ 1s and $2.4.5=40$ 2s.
answered Dec 7 '17 at 15:12
S. Dolan
1,205112
I don't see why $F(m)$ is congruent to $a!cdot (p-1)!^b F(b)$. Take, for example, $p=3, a=0,b=3$ so $m=9$. Then $k=1$ so $F(m)cong 1pmod 2$ but the formula you present gives $2$.
– Levent
Dec 9 '17 at 21:59
add a comment |
Same about
$F(m,p) = F(m mod p , p) dot F(m div p , p) dot (-1)^{m div p} $
with the integer division.
answered Dec 10 '17 at 0:41
kotomord
1,460626
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Does not seem to be that simple as we let things run high; I did primes 5, 7, 11. Seems reasonable to suggest 1, p-1 are the same and highest, after that not clear, although 5, 7 seem symmetric. I ran 11 pretty high, I guess if a+b = 11 then their counts are similar.
1 count 1330 2 count 1169 3 count 1169 4 count 1332
1 count 2570 2 count 2220 3 count 2230 4 count 2210 5 count 2220 6 count 2550
1 count 121894 2 count 115087 3 count 99710 4 count 100842 5 count 113429
6 count 112909 7 count 99693 8 count 99050 9 count 114123 10 count 123263
int main()
{
int p = 11;
int count[p] ;
int fac = 1;
for(int i = 1; i < p; ++i) count[i] = 0;
for(int i = 1; i <= 100000 * p; ++i)
{
int n = i;
while ( n % p == 0 ) n /= p;
n %= p;
fac *= n;
fac %= p;
count[fac] = 1 + count[fac];
cout << i << " " << fac << " count " << count[fac] << endl;
}
for(int i = 1; i < p; ++i) cout << i << " count " << count[i] << " " ;
cout << endl << endl;
return 0 ;
}
answered Jul 6 '16 at 22:15
Will Jagy
101k599199
1
I had similar results on those primes, but on 13, 17, 19 a result of "1" stayed high but the remaining values started to flatten out (testing to 30000)
– Joffan
Jul 6 '16 at 22:38
add a comment |
Does not seem to be that simple as we let things run high; I did primes 5, 7, 11. Seems reasonable to suggest 1, p-1 are the same and highest, after that not clear, although 5, 7 seem symmetric. I ran 11 pretty high, I guess if a+b = 11 then their counts are similar.
1 count 1330 2 count 1169 3 count 1169 4 count 1332
1 count 2570 2 count 2220 3 count 2230 4 count 2210 5 count 2220 6 count 2550
1 count 121894 2 count 115087 3 count 99710 4 count 100842 5 count 113429
6 count 112909 7 count 99693 8 count 99050 9 count 114123 10 count 123263
int main()
{
int p = 11;
int count[p] ;
int fac = 1;
for(int i = 1; i < p; ++i) count[i] = 0;
for(int i = 1; i <= 100000 * p; ++i)
{
int n = i;
while ( n % p == 0 ) n /= p;
n %= p;
fac *= n;
fac %= p;
count[fac] = 1 + count[fac];
cout << i << " " << fac << " count " << count[fac] << endl;
}
for(int i = 1; i < p; ++i) cout << i << " count " << count[i] << " " ;
cout << endl << endl;
return 0 ;
}
answered Jul 6 '16 at 22:15
Will Jagy
101k599199
1
I had similar results on those primes, but on 13, 17, 19 a result of "1" stayed high but the remaining values started to flatten out (testing to 30000)
– Joffan
Jul 6 '16 at 22:38
add a comment |
Does not seem to be that simple as we let things run high; I did primes 5, 7, 11. Seems reasonable to suggest 1, p-1 are the same and highest, after that not clear, although 5, 7 seem symmetric. I ran 11 pretty high, I guess if a+b = 11 then their counts are similar.
1 count 1330 2 count 1169 3 count 1169 4 count 1332
1 count 2570 2 count 2220 3 count 2230 4 count 2210 5 count 2220 6 count 2550
1 count 121894 2 count 115087 3 count 99710 4 count 100842 5 count 113429
6 count 112909 7 count 99693 8 count 99050 9 count 114123 10 count 123263
int main()
{
int p = 11;
int count[p] ;
int fac = 1;
for(int i = 1; i < p; ++i) count[i] = 0;
for(int i = 1; i <= 100000 * p; ++i)
{
int n = i;
while ( n % p == 0 ) n /= p;
n %= p;
fac *= n;
fac %= p;
count[fac] = 1 + count[fac];
cout << i << " " << fac << " count " << count[fac] << endl;
}
for(int i = 1; i < p; ++i) cout << i << " count " << count[i] << " " ;
cout << endl << endl;
return 0 ;
}
answered Jul 6 '16 at 22:15
Will Jagy
101k599199
Does not seem to be that simple as we let things run high; I did primes 5, 7, 11. Seems reasonable to suggest 1, p-1 are the same and highest, after that not clear, although 5, 7 seem symmetric. I ran 11 pretty high, I guess if a+b = 11 then their counts are similar.
1 count 1330 2 count 1169 3 count 1169 4 count 1332
1 count 2570 2 count 2220 3 count 2230 4 count 2210 5 count 2220 6 count 2550
1 count 121894 2 count 115087 3 count 99710 4 count 100842 5 count 113429
6 count 112909 7 count 99693 8 count 99050 9 count 114123 10 count 123263
int main()
{
int p = 11;
int count[p] ;
int fac = 1;
for(int i = 1; i < p; ++i) count[i] = 0;
for(int i = 1; i <= 100000 * p; ++i)
{
int n = i;
while ( n % p == 0 ) n /= p;
n %= p;
fac *= n;
fac %= p;
count[fac] = 1 + count[fac];
cout << i << " " << fac << " count " << count[fac] << endl;
}
for(int i = 1; i < p; ++i) cout << i << " count " << count[i] << " " ;
cout << endl << endl;
return 0 ;
}
answered Jul 6 '16 at 22:15
Will Jagy
101k599199
answered Jul 6 '16 at 22:15
Will Jagy
101k599199
answered Jul 6 '16 at 22:15
Will Jagy
101k599199
answered Jul 6 '16 at 22:15
Will Jagy
101k599199
101k599199
1
I had similar results on those primes, but on 13, 17, 19 a result of "1" stayed high but the remaining values started to flatten out (testing to 30000)
– Joffan
Jul 6 '16 at 22:38
add a comment |
1
I had similar results on those primes, but on 13, 17, 19 a result of "1" stayed high but the remaining values started to flatten out (testing to 30000)
– Joffan
Jul 6 '16 at 22:38
1
1
I had similar results on those primes, but on 13, 17, 19 a result of "1" stayed high but the remaining values started to flatten out (testing to 30000)
– Joffan
Jul 6 '16 at 22:38
I had similar results on those primes, but on 13, 17, 19 a result of "1" stayed high but the remaining values started to flatten out (testing to 30000)
– Joffan
Jul 6 '16 at 22:38
add a comment |
With $k$ as in the question, define $F(m)$ to be the residue of $k$ modulo $p$.
A formula
Let $m=a+bp$, where $0le ale p-1$. Then, modulo $p$, $F(m)$ is congruent to $a!(p-1)!^bF(b)$. By Wilson's Theorem this simplifies to $a!(-1)^bF(b)$.
So, in general, let $$m=a_0+a_1p+a_2p^2+...+a_{2n-1}p^{2n-1},0le a_ile p-1.$$
Then, for $A=a_1+a_3+...+a_{2n-1}$, $F(m)$ is congruent, modulo $p$, to
$$a_0!a_1!...a_{2n-1}!(-1)^A.$$
Frequencies
The formula shows that $F(m)$ is the product modulo $p$ $$F(a_0+a_1p) F(a_2+a_3p)...F(a_{2n-2}+a_{2n-1}p).$$
Therefore the frequencies for the $p^2$numbers $0!,1!,...,(p^2-1)!$ determine the frequencies for the $p^4$numbers up to $(p^4-1)!$ etc. etc.
For example, modulo 3 there are 4 1s and 5 2s from 0! to 8!
So up to 80! there are $4^2+5^2=41$ 1s and $2.4.5=40$ 2s.
answered Dec 7 '17 at 15:12
S. Dolan
1,205112
I don't see why $F(m)$ is congruent to $a!cdot (p-1)!^b F(b)$. Take, for example, $p=3, a=0,b=3$ so $m=9$. Then $k=1$ so $F(m)cong 1pmod 2$ but the formula you present gives $2$.
– Levent
Dec 9 '17 at 21:59
add a comment |
With $k$ as in the question, define $F(m)$ to be the residue of $k$ modulo $p$.
A formula
Let $m=a+bp$, where $0le ale p-1$. Then, modulo $p$, $F(m)$ is congruent to $a!(p-1)!^bF(b)$. By Wilson's Theorem this simplifies to $a!(-1)^bF(b)$.
So, in general, let $$m=a_0+a_1p+a_2p^2+...+a_{2n-1}p^{2n-1},0le a_ile p-1.$$
Then, for $A=a_1+a_3+...+a_{2n-1}$, $F(m)$ is congruent, modulo $p$, to
$$a_0!a_1!...a_{2n-1}!(-1)^A.$$
Frequencies
The formula shows that $F(m)$ is the product modulo $p$ $$F(a_0+a_1p) F(a_2+a_3p)...F(a_{2n-2}+a_{2n-1}p).$$
Therefore the frequencies for the $p^2$numbers $0!,1!,...,(p^2-1)!$ determine the frequencies for the $p^4$numbers up to $(p^4-1)!$ etc. etc.
For example, modulo 3 there are 4 1s and 5 2s from 0! to 8!
So up to 80! there are $4^2+5^2=41$ 1s and $2.4.5=40$ 2s.
answered Dec 7 '17 at 15:12
S. Dolan
1,205112
I don't see why $F(m)$ is congruent to $a!cdot (p-1)!^b F(b)$. Take, for example, $p=3, a=0,b=3$ so $m=9$. Then $k=1$ so $F(m)cong 1pmod 2$ but the formula you present gives $2$.
– Levent
Dec 9 '17 at 21:59
add a comment |
With $k$ as in the question, define $F(m)$ to be the residue of $k$ modulo $p$.
A formula
Let $m=a+bp$, where $0le ale p-1$. Then, modulo $p$, $F(m)$ is congruent to $a!(p-1)!^bF(b)$. By Wilson's Theorem this simplifies to $a!(-1)^bF(b)$.
So, in general, let $$m=a_0+a_1p+a_2p^2+...+a_{2n-1}p^{2n-1},0le a_ile p-1.$$
Then, for $A=a_1+a_3+...+a_{2n-1}$, $F(m)$ is congruent, modulo $p$, to
$$a_0!a_1!...a_{2n-1}!(-1)^A.$$
Frequencies
The formula shows that $F(m)$ is the product modulo $p$ $$F(a_0+a_1p) F(a_2+a_3p)...F(a_{2n-2}+a_{2n-1}p).$$
Therefore the frequencies for the $p^2$numbers $0!,1!,...,(p^2-1)!$ determine the frequencies for the $p^4$numbers up to $(p^4-1)!$ etc. etc.
For example, modulo 3 there are 4 1s and 5 2s from 0! to 8!
So up to 80! there are $4^2+5^2=41$ 1s and $2.4.5=40$ 2s.
answered Dec 7 '17 at 15:12
S. Dolan
1,205112
With $k$ as in the question, define $F(m)$ to be the residue of $k$ modulo $p$.
A formula
Let $m=a+bp$, where $0le ale p-1$. Then, modulo $p$, $F(m)$ is congruent to $a!(p-1)!^bF(b)$. By Wilson's Theorem this simplifies to $a!(-1)^bF(b)$.
So, in general, let $$m=a_0+a_1p+a_2p^2+...+a_{2n-1}p^{2n-1},0le a_ile p-1.$$
Then, for $A=a_1+a_3+...+a_{2n-1}$, $F(m)$ is congruent, modulo $p$, to
$$a_0!a_1!...a_{2n-1}!(-1)^A.$$
Frequencies
The formula shows that $F(m)$ is the product modulo $p$ $$F(a_0+a_1p) F(a_2+a_3p)...F(a_{2n-2}+a_{2n-1}p).$$
Therefore the frequencies for the $p^2$numbers $0!,1!,...,(p^2-1)!$ determine the frequencies for the $p^4$numbers up to $(p^4-1)!$ etc. etc.
For example, modulo 3 there are 4 1s and 5 2s from 0! to 8!
So up to 80! there are $4^2+5^2=41$ 1s and $2.4.5=40$ 2s.
answered Dec 7 '17 at 15:12
S. Dolan
1,205112
answered Dec 7 '17 at 15:12
S. Dolan
1,205112
answered Dec 7 '17 at 15:12
S. Dolan
1,205112
answered Dec 7 '17 at 15:12
S. Dolan
1,205112
1,205112
I don't see why $F(m)$ is congruent to $a!cdot (p-1)!^b F(b)$. Take, for example, $p=3, a=0,b=3$ so $m=9$. Then $k=1$ so $F(m)cong 1pmod 2$ but the formula you present gives $2$.
– Levent
Dec 9 '17 at 21:59
add a comment |
I don't see why $F(m)$ is congruent to $a!cdot (p-1)!^b F(b)$. Take, for example, $p=3, a=0,b=3$ so $m=9$. Then $k=1$ so $F(m)cong 1pmod 2$ but the formula you present gives $2$.
– Levent
Dec 9 '17 at 21:59
I don't see why $F(m)$ is congruent to $a!cdot (p-1)!^b F(b)$. Take, for example, $p=3, a=0,b=3$ so $m=9$. Then $k=1$ so $F(m)cong 1pmod 2$ but the formula you present gives $2$.
– Levent
Dec 9 '17 at 21:59
I don't see why $F(m)$ is congruent to $a!cdot (p-1)!^b F(b)$. Take, for example, $p=3, a=0,b=3$ so $m=9$. Then $k=1$ so $F(m)cong 1pmod 2$ but the formula you present gives $2$.
– Levent
Dec 9 '17 at 21:59
add a comment |
Same about
$F(m,p) = F(m mod p , p) dot F(m div p , p) dot (-1)^{m div p} $
with the integer division.
answered Dec 10 '17 at 0:41
kotomord
1,460626
add a comment |
Same about
$F(m,p) = F(m mod p , p) dot F(m div p , p) dot (-1)^{m div p} $
with the integer division.
answered Dec 10 '17 at 0:41
kotomord
1,460626
add a comment |
Same about
$F(m,p) = F(m mod p , p) dot F(m div p , p) dot (-1)^{m div p} $
with the integer division.
answered Dec 10 '17 at 0:41
kotomord
1,460626
Same about
$F(m,p) = F(m mod p , p) dot F(m div p , p) dot (-1)^{m div p} $
with the integer division.
answered Dec 10 '17 at 0:41
kotomord
1,460626
answered Dec 10 '17 at 0:41
kotomord
1,460626
answered Dec 10 '17 at 0:41
kotomord
1,460626
answered Dec 10 '17 at 0:41
kotomord
1,460626
1,460626
add a comment |
add a comment |
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"frequencies" in what sense? across all higher factorials?
– Joffan
Jul 6 '16 at 21:50
I simply meant when I take $m=50$ and $p=3,5$ or $7$, the number of occurrences of $1,2,...,p-1$ as remainders are almost equal.
– Levent
Jul 6 '16 at 21:54
You can't talk about a frequency when you only look at one value of $m$.
– Joffan
Jul 6 '16 at 22:42
Sorry, what I mean is when I take $m=2$ up to $50$.
– Levent
Jul 6 '16 at 22:51