Using a Do-loop to find divisors mod 13












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







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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









    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.







    $endgroup$















      1












      1








      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









      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.







      $endgroup$




      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




      core-language number-theory






      share|improve this question









      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









      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




      share|improve this question








      edited 5 hours ago









      m_goldberg

      86.4k872196




      86.4k872196






      New contributor




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      asked 7 hours ago









      argamonargamon

      82




      82




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      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.






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






















          2 Answers
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          $begingroup$

          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











          $endgroup$













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










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



















          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









          $endgroup$













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











          Your Answer





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          2 Answers
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          active

          oldest

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          2 Answers
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          active

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          active

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          active

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          $begingroup$

          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











          $endgroup$













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










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
















          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











          $endgroup$













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










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














          2












          2








          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











          $endgroup$



          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}*)


          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














          share|improve this answer



          share|improve this answer








          edited 6 hours ago

























          answered 6 hours ago









          Bill WattsBill Watts

          3,4311620




          3,4311620












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










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


















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










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
















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




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












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




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











          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









          $endgroup$













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
















          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









          $endgroup$













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














          2












          2








          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









          $endgroup$



          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












          share|improve this answer



          share|improve this answer










          answered 6 hours ago









          kglrkglr

          185k10202420




          185k10202420












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


















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
















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




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










          argamon is a new contributor. Be nice, and check out our Code of Conduct.










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