a question about the notation in the book “Opera de Cribro”
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When I study the book "Opera De Cribro" by John Friedlander, Henryk Iwaniec-(2010), in Sections 1.2 and 1.3, I confused with notation used there. In page 3 it is defined:
$$cal{A} = (a_n) , nle x$$,
$$A(x)=sum_{nle x} a_n$$
$$cal{A}_d=(am) , m equiv 0 (mod d)$$
and
$$A_d(x) =sum_{substack{nle x\ nequiv 0 (mod d)}} a_n$$
Now my question happens in the example 1.1 on page 5.
what are $a_n$ here? I wonder if $a_n=1$ or $a_n=n$?
Thanks.
number-theory elementary-number-theory notation analytic-number-theory sieve-theory
asked Dec 10 '18 at 11:13
asadasad
36719
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add a comment |
$begingroup$
When I study the book "Opera De Cribro" by John Friedlander, Henryk Iwaniec-(2010), in Sections 1.2 and 1.3, I confused with notation used there. In page 3 it is defined:
$$cal{A} = (a_n) , nle x$$,
$$A(x)=sum_{nle x} a_n$$
$$cal{A}_d=(am) , m equiv 0 (mod d)$$
and
$$A_d(x) =sum_{substack{nle x\ nequiv 0 (mod d)}} a_n$$
Now my question happens in the example 1.1 on page 5.
what are $a_n$ here? I wonder if $a_n=1$ or $a_n=n$?
Thanks.
number-theory elementary-number-theory notation analytic-number-theory sieve-theory
asked Dec 10 '18 at 11:13
asadasad
36719
$endgroup$
$begingroup$
It is $a_n = 1$ for $n in (x-y,x]$, $a_n = 0$ otherwise, and $A_d$ is based on $b_n = 1$ for $n in (x-y,x], d | n$, $b_n = 0$ otherwise
$endgroup$
– reuns
Dec 10 '18 at 14:08
1
$begingroup$
Crossposted at MO: mathoverflow.net/questions/317332/…
$endgroup$
– Noah Schweber
Dec 10 '18 at 15:09
add a comment |
$begingroup$
When I study the book "Opera De Cribro" by John Friedlander, Henryk Iwaniec-(2010), in Sections 1.2 and 1.3, I confused with notation used there. In page 3 it is defined:
$$cal{A} = (a_n) , nle x$$,
$$A(x)=sum_{nle x} a_n$$
$$cal{A}_d=(am) , m equiv 0 (mod d)$$
and
$$A_d(x) =sum_{substack{nle x\ nequiv 0 (mod d)}} a_n$$
Now my question happens in the example 1.1 on page 5.
what are $a_n$ here? I wonder if $a_n=1$ or $a_n=n$?
Thanks.
number-theory elementary-number-theory notation analytic-number-theory sieve-theory
asked Dec 10 '18 at 11:13
asadasad
36719
$endgroup$
When I study the book "Opera De Cribro" by John Friedlander, Henryk Iwaniec-(2010), in Sections 1.2 and 1.3, I confused with notation used there. In page 3 it is defined:
$$cal{A} = (a_n) , nle x$$,
$$A(x)=sum_{nle x} a_n$$
$$cal{A}_d=(am) , m equiv 0 (mod d)$$
and
$$A_d(x) =sum_{substack{nle x\ nequiv 0 (mod d)}} a_n$$
Now my question happens in the example 1.1 on page 5.
what are $a_n$ here? I wonder if $a_n=1$ or $a_n=n$?
Thanks.
number-theory elementary-number-theory notation analytic-number-theory sieve-theory
number-theory elementary-number-theory notation analytic-number-theory sieve-theory
asked Dec 10 '18 at 11:13
asadasad
36719
asked Dec 10 '18 at 11:13
asadasad
36719
edited Dec 10 '18 at 11:36
asad
asked Dec 10 '18 at 11:13
asadasad
36719
asked Dec 10 '18 at 11:13
asadasad
36719
asked Dec 10 '18 at 11:13
asadasad
36719
36719
$begingroup$
It is $a_n = 1$ for $n in (x-y,x]$, $a_n = 0$ otherwise, and $A_d$ is based on $b_n = 1$ for $n in (x-y,x], d | n$, $b_n = 0$ otherwise
$endgroup$
– reuns
Dec 10 '18 at 14:08
1
$begingroup$
Crossposted at MO: mathoverflow.net/questions/317332/…
$endgroup$
– Noah Schweber
Dec 10 '18 at 15:09
add a comment |
$begingroup$
It is $a_n = 1$ for $n in (x-y,x]$, $a_n = 0$ otherwise, and $A_d$ is based on $b_n = 1$ for $n in (x-y,x], d | n$, $b_n = 0$ otherwise
$endgroup$
– reuns
Dec 10 '18 at 14:08
1
$begingroup$
Crossposted at MO: mathoverflow.net/questions/317332/…
$endgroup$
– Noah Schweber
Dec 10 '18 at 15:09
$begingroup$
It is $a_n = 1$ for $n in (x-y,x]$, $a_n = 0$ otherwise, and $A_d$ is based on $b_n = 1$ for $n in (x-y,x], d | n$, $b_n = 0$ otherwise
$endgroup$
– reuns
Dec 10 '18 at 14:08
$begingroup$
It is $a_n = 1$ for $n in (x-y,x]$, $a_n = 0$ otherwise, and $A_d$ is based on $b_n = 1$ for $n in (x-y,x], d | n$, $b_n = 0$ otherwise
$endgroup$
– reuns
Dec 10 '18 at 14:08
1
1
$begingroup$
Crossposted at MO: mathoverflow.net/questions/317332/…
$endgroup$
– Noah Schweber
Dec 10 '18 at 15:09
$begingroup$
Crossposted at MO: mathoverflow.net/questions/317332/…
$endgroup$
– Noah Schweber
Dec 10 '18 at 15:09
add a comment |
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$begingroup$
It is $a_n = 1$ for $n in (x-y,x]$, $a_n = 0$ otherwise, and $A_d$ is based on $b_n = 1$ for $n in (x-y,x], d | n$, $b_n = 0$ otherwise
$endgroup$
– reuns
Dec 10 '18 at 14:08
1
$begingroup$
Crossposted at MO: mathoverflow.net/questions/317332/…
$endgroup$
– Noah Schweber
Dec 10 '18 at 15:09