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Using a Do-loop to find divisors mod 13

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1

$begingroup$

I want to check sum of divisors of i mod 13 fori = 1 to i = 20. I tried writing a Do-Print loop so it looks like

Do[Print[DivisorSigma[1, i], {i, 20} mod 13] 

Any help will be greatly appreciated,

core-language number-theory

share|improve this question

edited 5 hours ago

m_goldberg

86.4k872196

asked 7 hours ago

argamonargamon

82

New contributor

argamon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

add a comment | 

1

$begingroup$

I want to check sum of divisors of i mod 13 fori = 1 to i = 20. I tried writing a Do-Print loop so it looks like

Do[Print[DivisorSigma[1, i], {i, 20} mod 13] 

Any help will be greatly appreciated,

core-language number-theory

share|improve this question

edited 5 hours ago

m_goldberg

86.4k872196

asked 7 hours ago

argamonargamon

82

New contributor

argamon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

I want to check sum of divisors of i mod 13 fori = 1 to i = 20. I tried writing a Do-Print loop so it looks like

Do[Print[DivisorSigma[1, i], {i, 20} mod 13] 

Any help will be greatly appreciated,

core-language number-theory

share|improve this question

edited 5 hours ago

m_goldberg

86.4k872196

asked 7 hours ago

argamonargamon

82

New contributor

argamon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

Do[Print[DivisorSigma[1, i], {i, 20} mod 13] 

share|improve this question

edited 5 hours ago

m_goldberg

86.4k872196

asked 7 hours ago

argamonargamon

82

New contributor

argamon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 5 hours ago

m_goldberg

86.4k872196

asked 7 hours ago

argamonargamon

82

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argamon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 5 hours ago

m_goldberg

86.4k872196

edited 5 hours ago

m_goldberg

86.4k872196

asked 7 hours ago

argamonargamon

82

New contributor

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asked 7 hours ago

argamonargamon

82

2 Answers
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2

$begingroup$

Table[DivisorSigma[1, i], {i, 20} ] // Mod[#, 13] &

(*{1, 3, 4, 7, 6, 12, 8, 2, 0, 5, 12, 2, 1, 11, 11, 5, 5, 0, 7, 3}*)

or

Do[Print[Mod[DivisorSigma[1, i], 13]], {i, 20}]

which gives a column of the same numbers as before that you can look at, but it is more difficult use in further expressions.

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Bill WattsBill Watts

3,4311620

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you so much
  $endgroup$
  – argamon
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome.
  $endgroup$
  – Bill Watts
  6 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

You can do it without iterators because the functions you need (Mod, DivisorSigma) are Listable:

Mod[DivisorSigma[1, Range[20]], 13]

{1, 3, 4, 7, 6, 12, 8, 2, 0, 5, 12, 2, 1, 11, 11, 5, 5, 0, 7, 3}

Note: Listable functions are applied separately to each element in a list.

share|improve this answer

answered 6 hours ago

kglrkglr

185k10202420

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  that's also very useful and helpful thank you
  $endgroup$
  – argamon
  6 hours ago
  

  
  
  
  

add a comment | 

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2

$begingroup$

Table[DivisorSigma[1, i], {i, 20} ] // Mod[#, 13] &

(*{1, 3, 4, 7, 6, 12, 8, 2, 0, 5, 12, 2, 1, 11, 11, 5, 5, 0, 7, 3}*)

or

Do[Print[Mod[DivisorSigma[1, i], 13]], {i, 20}]

which gives a column of the same numbers as before that you can look at, but it is more difficult use in further expressions.

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Bill WattsBill Watts

3,4311620

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you so much
  $endgroup$
  – argamon
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome.
  $endgroup$
  – Bill Watts
  6 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Table[DivisorSigma[1, i], {i, 20} ] // Mod[#, 13] &

(*{1, 3, 4, 7, 6, 12, 8, 2, 0, 5, 12, 2, 1, 11, 11, 5, 5, 0, 7, 3}*)

or

Do[Print[Mod[DivisorSigma[1, i], 13]], {i, 20}]

which gives a column of the same numbers as before that you can look at, but it is more difficult use in further expressions.

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Bill WattsBill Watts

3,4311620

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you so much
  $endgroup$
  – argamon
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome.
  $endgroup$
  – Bill Watts
  6 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Table[DivisorSigma[1, i], {i, 20} ] // Mod[#, 13] &

(*{1, 3, 4, 7, 6, 12, 8, 2, 0, 5, 12, 2, 1, 11, 11, 5, 5, 0, 7, 3}*)

or

Do[Print[Mod[DivisorSigma[1, i], 13]], {i, 20}]

which gives a column of the same numbers as before that you can look at, but it is more difficult use in further expressions.

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Bill WattsBill Watts

3,4311620

$endgroup$

Table[DivisorSigma[1, i], {i, 20} ] // Mod[#, 13] &

(*{1, 3, 4, 7, 6, 12, 8, 2, 0, 5, 12, 2, 1, 11, 11, 5, 5, 0, 7, 3}*)

Do[Print[Mod[DivisorSigma[1, i], 13]], {i, 20}]

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Bill WattsBill Watts

3,4311620

answered 6 hours ago

Bill WattsBill Watts

3,4311620

answered 6 hours ago

Bill WattsBill Watts

3,4311620

2

$begingroup$

You can do it without iterators because the functions you need (Mod, DivisorSigma) are Listable:

Mod[DivisorSigma[1, Range[20]], 13]

{1, 3, 4, 7, 6, 12, 8, 2, 0, 5, 12, 2, 1, 11, 11, 5, 5, 0, 7, 3}

Note: Listable functions are applied separately to each element in a list.

share|improve this answer

answered 6 hours ago

kglrkglr

185k10202420

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  that's also very useful and helpful thank you
  $endgroup$
  – argamon
  6 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

You can do it without iterators because the functions you need (Mod, DivisorSigma) are Listable:

Mod[DivisorSigma[1, Range[20]], 13]

{1, 3, 4, 7, 6, 12, 8, 2, 0, 5, 12, 2, 1, 11, 11, 5, 5, 0, 7, 3}

Note: Listable functions are applied separately to each element in a list.

share|improve this answer

answered 6 hours ago

kglrkglr

185k10202420

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  that's also very useful and helpful thank you
  $endgroup$
  – argamon
  6 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

You can do it without iterators because the functions you need (Mod, DivisorSigma) are Listable:

Mod[DivisorSigma[1, Range[20]], 13]

{1, 3, 4, 7, 6, 12, 8, 2, 0, 5, 12, 2, 1, 11, 11, 5, 5, 0, 7, 3}

Note: Listable functions are applied separately to each element in a list.

share|improve this answer

answered 6 hours ago

kglrkglr

185k10202420

$endgroup$

Mod[DivisorSigma[1, Range[20]], 13]

share|improve this answer

answered 6 hours ago

kglrkglr

185k10202420

answered 6 hours ago

kglrkglr

185k10202420

answered 6 hours ago

kglrkglr

185k10202420