a question about the notation in the book “Opera de Cribro”












0












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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.



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number-theory elementary-number-theory notation analytic-number-theory sieve-theory







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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
















0












$begingroup$


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.



enter image description here






number-theory elementary-number-theory notation analytic-number-theory sieve-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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.



enter image description here






number-theory elementary-number-theory notation analytic-number-theory sieve-theory







share|cite|improve this question











$endgroup$




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.



enter image description here






number-theory elementary-number-theory notation analytic-number-theory sieve-theory




number-theory elementary-number-theory notation analytic-number-theory sieve-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 11:36







asad

















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


















  • $begingroup$
    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










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