Ramanujan's congruence.












2












$begingroup$


There are the following formula in this link(https://en.wikipedia.org/wiki/Ramanujan%27s_congruences).




$$sum_{k=0}^{ infty } p(5k+4)q^k = 5 frac{(q^5)_{infty}^{5}}{(q)_{infty}^{6}}.$$



$$sum_{k=0}^{ infty } p(7k+5)q^k = 7 frac{(q^7)_{infty}^{3}}{(q)_{infty}^{4}} + 49q frac{(q^7)_{infty}^{7}}{(q)_{infty}^{8}}.$$




Srinivasa Ramanujan discovered the following congruences.




$$ p(11k+6) = 0　mod.11.$$




Do you know the formla of $sum_{k=0}^{ infty } p(11k+6)q^k $?



That is




$$sum_{k=0}^{ infty } p(11k+6)q^k = ???.$$







number-theory integer-partitions







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    There are the following formula in this link(https://en.wikipedia.org/wiki/Ramanujan%27s_congruences).




    $$sum_{k=0}^{ infty } p(5k+4)q^k = 5 frac{(q^5)_{infty}^{5}}{(q)_{infty}^{6}}.$$



    $$sum_{k=0}^{ infty } p(7k+5)q^k = 7 frac{(q^7)_{infty}^{3}}{(q)_{infty}^{4}} + 49q frac{(q^7)_{infty}^{7}}{(q)_{infty}^{8}}.$$




    Srinivasa Ramanujan discovered the following congruences.




    $$ p(11k+6) = 0　mod.11.$$




    Do you know the formla of $sum_{k=0}^{ infty } p(11k+6)q^k $?



    That is




    $$sum_{k=0}^{ infty } p(11k+6)q^k = ???.$$







    number-theory integer-partitions







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      There are the following formula in this link(https://en.wikipedia.org/wiki/Ramanujan%27s_congruences).




      $$sum_{k=0}^{ infty } p(5k+4)q^k = 5 frac{(q^5)_{infty}^{5}}{(q)_{infty}^{6}}.$$



      $$sum_{k=0}^{ infty } p(7k+5)q^k = 7 frac{(q^7)_{infty}^{3}}{(q)_{infty}^{4}} + 49q frac{(q^7)_{infty}^{7}}{(q)_{infty}^{8}}.$$




      Srinivasa Ramanujan discovered the following congruences.




      $$ p(11k+6) = 0　mod.11.$$




      Do you know the formla of $sum_{k=0}^{ infty } p(11k+6)q^k $?



      That is




      $$sum_{k=0}^{ infty } p(11k+6)q^k = ???.$$







      number-theory integer-partitions







      share|cite|improve this question











      $endgroup$




      There are the following formula in this link(https://en.wikipedia.org/wiki/Ramanujan%27s_congruences).




      $$sum_{k=0}^{ infty } p(5k+4)q^k = 5 frac{(q^5)_{infty}^{5}}{(q)_{infty}^{6}}.$$



      $$sum_{k=0}^{ infty } p(7k+5)q^k = 7 frac{(q^7)_{infty}^{3}}{(q)_{infty}^{4}} + 49q frac{(q^7)_{infty}^{7}}{(q)_{infty}^{8}}.$$




      Srinivasa Ramanujan discovered the following congruences.




      $$ p(11k+6) = 0　mod.11.$$




      Do you know the formla of $sum_{k=0}^{ infty } p(11k+6)q^k $?



      That is




      $$sum_{k=0}^{ infty } p(11k+6)q^k = ???.$$







      number-theory integer-partitions




      number-theory integer-partitions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 13 '16 at 13:06







      Manyama

















      asked Nov 13 '16 at 11:59









      ManyamaManyama

      5021415




      5021415






















          1 Answer
          1






          active

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          2












          $begingroup$

          Yes, a formula for the series $sum_{k=0}^{ infty } p(11k+6)q^k $, written as $sum_{n=1}^{ infty } p(11n-5)q^n$, can be found in the paper Ramanujan's unpublished manuscript on the partition and tau functions with proofs and commentary by Bruce C. Berndt and Ken Ono, on page $39$, formula $(18.7)$:



          $$
          sum_{n=1}^{ infty } p(11n-5)q^n(q_{11};q_{11})_{infty}=11sigma_7(n)q^n+121J,
          $$
          which can be rewritten as you want, I suppose.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            In the case of 5 and 7, there are the formula in the paper(Those are (4.7) and (8.3)). Are there the formula in the case of not 121 but 11?
            $endgroup$
            – Manyama
            Nov 13 '16 at 13:01






          • 1




            $begingroup$
            Berndt and Ono would have noted further nice formulas of this type, I suppose. So perhaps it is more complicated.
            $endgroup$
            – Dietrich Burde
            Nov 13 '16 at 13:15












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          1 Answer
          1






          active

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          active

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          active

          oldest

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          2












          $begingroup$

          Yes, a formula for the series $sum_{k=0}^{ infty } p(11k+6)q^k $, written as $sum_{n=1}^{ infty } p(11n-5)q^n$, can be found in the paper Ramanujan's unpublished manuscript on the partition and tau functions with proofs and commentary by Bruce C. Berndt and Ken Ono, on page $39$, formula $(18.7)$:



          $$
          sum_{n=1}^{ infty } p(11n-5)q^n(q_{11};q_{11})_{infty}=11sigma_7(n)q^n+121J,
          $$
          which can be rewritten as you want, I suppose.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            In the case of 5 and 7, there are the formula in the paper(Those are (4.7) and (8.3)). Are there the formula in the case of not 121 but 11?
            $endgroup$
            – Manyama
            Nov 13 '16 at 13:01






          • 1




            $begingroup$
            Berndt and Ono would have noted further nice formulas of this type, I suppose. So perhaps it is more complicated.
            $endgroup$
            – Dietrich Burde
            Nov 13 '16 at 13:15
















          2












          $begingroup$

          Yes, a formula for the series $sum_{k=0}^{ infty } p(11k+6)q^k $, written as $sum_{n=1}^{ infty } p(11n-5)q^n$, can be found in the paper Ramanujan's unpublished manuscript on the partition and tau functions with proofs and commentary by Bruce C. Berndt and Ken Ono, on page $39$, formula $(18.7)$:



          $$
          sum_{n=1}^{ infty } p(11n-5)q^n(q_{11};q_{11})_{infty}=11sigma_7(n)q^n+121J,
          $$
          which can be rewritten as you want, I suppose.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            In the case of 5 and 7, there are the formula in the paper(Those are (4.7) and (8.3)). Are there the formula in the case of not 121 but 11?
            $endgroup$
            – Manyama
            Nov 13 '16 at 13:01






          • 1




            $begingroup$
            Berndt and Ono would have noted further nice formulas of this type, I suppose. So perhaps it is more complicated.
            $endgroup$
            – Dietrich Burde
            Nov 13 '16 at 13:15














          2












          2








          2





          $begingroup$

          Yes, a formula for the series $sum_{k=0}^{ infty } p(11k+6)q^k $, written as $sum_{n=1}^{ infty } p(11n-5)q^n$, can be found in the paper Ramanujan's unpublished manuscript on the partition and tau functions with proofs and commentary by Bruce C. Berndt and Ken Ono, on page $39$, formula $(18.7)$:



          $$
          sum_{n=1}^{ infty } p(11n-5)q^n(q_{11};q_{11})_{infty}=11sigma_7(n)q^n+121J,
          $$
          which can be rewritten as you want, I suppose.






          share|cite|improve this answer











          $endgroup$



          Yes, a formula for the series $sum_{k=0}^{ infty } p(11k+6)q^k $, written as $sum_{n=1}^{ infty } p(11n-5)q^n$, can be found in the paper Ramanujan's unpublished manuscript on the partition and tau functions with proofs and commentary by Bruce C. Berndt and Ken Ono, on page $39$, formula $(18.7)$:



          $$
          sum_{n=1}^{ infty } p(11n-5)q^n(q_{11};q_{11})_{infty}=11sigma_7(n)q^n+121J,
          $$
          which can be rewritten as you want, I suppose.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 29 at 10:01









          MichalisN

          4,6071327




          4,6071327










          answered Nov 13 '16 at 12:30









          Dietrich BurdeDietrich Burde

          81.6k648106




          81.6k648106












          • $begingroup$
            In the case of 5 and 7, there are the formula in the paper(Those are (4.7) and (8.3)). Are there the formula in the case of not 121 but 11?
            $endgroup$
            – Manyama
            Nov 13 '16 at 13:01






          • 1




            $begingroup$
            Berndt and Ono would have noted further nice formulas of this type, I suppose. So perhaps it is more complicated.
            $endgroup$
            – Dietrich Burde
            Nov 13 '16 at 13:15


















          • $begingroup$
            In the case of 5 and 7, there are the formula in the paper(Those are (4.7) and (8.3)). Are there the formula in the case of not 121 but 11?
            $endgroup$
            – Manyama
            Nov 13 '16 at 13:01






          • 1




            $begingroup$
            Berndt and Ono would have noted further nice formulas of this type, I suppose. So perhaps it is more complicated.
            $endgroup$
            – Dietrich Burde
            Nov 13 '16 at 13:15
















          $begingroup$
          In the case of 5 and 7, there are the formula in the paper(Those are (4.7) and (8.3)). Are there the formula in the case of not 121 but 11?
          $endgroup$
          – Manyama
          Nov 13 '16 at 13:01




          $begingroup$
          In the case of 5 and 7, there are the formula in the paper(Those are (4.7) and (8.3)). Are there the formula in the case of not 121 but 11?
          $endgroup$
          – Manyama
          Nov 13 '16 at 13:01




          1




          1




          $begingroup$
          Berndt and Ono would have noted further nice formulas of this type, I suppose. So perhaps it is more complicated.
          $endgroup$
          – Dietrich Burde
          Nov 13 '16 at 13:15




          $begingroup$
          Berndt and Ono would have noted further nice formulas of this type, I suppose. So perhaps it is more complicated.
          $endgroup$
          – Dietrich Burde
          Nov 13 '16 at 13:15


















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