Number of permutations that are products of disjoint cycles of distinct length

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What is the number of permutations $piin S_n$ that are products of disjoint cycles of distinct length? What is the number of permutations that are products of disjoint cycles such that no more than $k$ cycles are of any given length?



I am really interested in the asymptotics of the proportion of permutations in $S_n$ with these properties as $nto infty$. What is a good reference for this sort of statistics?










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  • Dumb question (from me): you are ignoring cycles of length one (fixed points), right? In which case, a good start on estimating the size of the complement is counting all permutations with two cycles of length two, and over counting by adding those with two cycles of length three. For back of the envelope estimates I would start with that. Gerhard "Higher Order Terms Can Wait" Paseman, 2018.11.02.
    – Gerhard Paseman
    2 hours ago











  • Actually, both versions of the question interest me - ignoring fixed points and considering them as cycles of length 1.
    – H A Helfgott
    1 hour ago














up vote
3
down vote

favorite












What is the number of permutations $piin S_n$ that are products of disjoint cycles of distinct length? What is the number of permutations that are products of disjoint cycles such that no more than $k$ cycles are of any given length?



I am really interested in the asymptotics of the proportion of permutations in $S_n$ with these properties as $nto infty$. What is a good reference for this sort of statistics?










share|cite|improve this question





















  • Dumb question (from me): you are ignoring cycles of length one (fixed points), right? In which case, a good start on estimating the size of the complement is counting all permutations with two cycles of length two, and over counting by adding those with two cycles of length three. For back of the envelope estimates I would start with that. Gerhard "Higher Order Terms Can Wait" Paseman, 2018.11.02.
    – Gerhard Paseman
    2 hours ago











  • Actually, both versions of the question interest me - ignoring fixed points and considering them as cycles of length 1.
    – H A Helfgott
    1 hour ago












up vote
3
down vote

favorite









up vote
3
down vote

favorite











What is the number of permutations $piin S_n$ that are products of disjoint cycles of distinct length? What is the number of permutations that are products of disjoint cycles such that no more than $k$ cycles are of any given length?



I am really interested in the asymptotics of the proportion of permutations in $S_n$ with these properties as $nto infty$. What is a good reference for this sort of statistics?










share|cite|improve this question













What is the number of permutations $piin S_n$ that are products of disjoint cycles of distinct length? What is the number of permutations that are products of disjoint cycles such that no more than $k$ cycles are of any given length?



I am really interested in the asymptotics of the proportion of permutations in $S_n$ with these properties as $nto infty$. What is a good reference for this sort of statistics?







permutations permutation-groups random-permutations






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asked 4 hours ago









H A Helfgott

4,42212579




4,42212579











  • Dumb question (from me): you are ignoring cycles of length one (fixed points), right? In which case, a good start on estimating the size of the complement is counting all permutations with two cycles of length two, and over counting by adding those with two cycles of length three. For back of the envelope estimates I would start with that. Gerhard "Higher Order Terms Can Wait" Paseman, 2018.11.02.
    – Gerhard Paseman
    2 hours ago











  • Actually, both versions of the question interest me - ignoring fixed points and considering them as cycles of length 1.
    – H A Helfgott
    1 hour ago
















  • Dumb question (from me): you are ignoring cycles of length one (fixed points), right? In which case, a good start on estimating the size of the complement is counting all permutations with two cycles of length two, and over counting by adding those with two cycles of length three. For back of the envelope estimates I would start with that. Gerhard "Higher Order Terms Can Wait" Paseman, 2018.11.02.
    – Gerhard Paseman
    2 hours ago











  • Actually, both versions of the question interest me - ignoring fixed points and considering them as cycles of length 1.
    – H A Helfgott
    1 hour ago















Dumb question (from me): you are ignoring cycles of length one (fixed points), right? In which case, a good start on estimating the size of the complement is counting all permutations with two cycles of length two, and over counting by adding those with two cycles of length three. For back of the envelope estimates I would start with that. Gerhard "Higher Order Terms Can Wait" Paseman, 2018.11.02.
– Gerhard Paseman
2 hours ago





Dumb question (from me): you are ignoring cycles of length one (fixed points), right? In which case, a good start on estimating the size of the complement is counting all permutations with two cycles of length two, and over counting by adding those with two cycles of length three. For back of the envelope estimates I would start with that. Gerhard "Higher Order Terms Can Wait" Paseman, 2018.11.02.
– Gerhard Paseman
2 hours ago













Actually, both versions of the question interest me - ignoring fixed points and considering them as cycles of length 1.
– H A Helfgott
1 hour ago




Actually, both versions of the question interest me - ignoring fixed points and considering them as cycles of length 1.
– H A Helfgott
1 hour ago










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If we denote by $c_i(sigma)$ the number of cycles of length $i$ in $sigma$, we can write the exponential generating function of permutations with cycle statistics as
$$sum_ngeq 1sum_sigmain S_nleft(fracx^nn!prod_igeq 1t_i^c_i(sigma)right)=expleft(sum_igeq 1 fract_ix^iiright) =prod_igeq 1 left(1+fract_ix^ii+fract_i^2x^2i2i^2+cdotsright)$$
From here we see that the exponential generating function of permutations with distinct cycle sizes can be obtained by removing all terms where any $t_i$ has exponent $geq 2$, and then setting all $t_i=1$. So we get
$$prod_igeq 1left(1+fracx^iiright)$$
From here the methods of A Hybrid of Darboux's Method and Singularity Analysis in Combinatorial Asymptotics by P. Flajolet, E. Fusy, X. Gourdon, D. Panario, N. Pouyanne show that the coefficient of $x^n$ is asymptotically equal to
$$e^-gamma+frace^-gamman+Oleft(fraclog nn^2right)$$
which means that our desired number of permutations is asymptotically given by $$n!left(e^-gamma+frace^-gamman+Oleft(fraclog nn^2right)right).$$
Notice that in section 3 they actually provide much more refined asymptotics, in case you wanted more terms. Moreover, I believe their method should let you compute the asymptotics for permutations with cycles of the same length appearing at most $k$ times whose generating function is given by
$$prod_igeq 1left(1+fracx^ii+fracx^2i2i^2+cdots +fracx^kik!i^kright)$$
from the same considerations as above.






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    If we denote by $c_i(sigma)$ the number of cycles of length $i$ in $sigma$, we can write the exponential generating function of permutations with cycle statistics as
    $$sum_ngeq 1sum_sigmain S_nleft(fracx^nn!prod_igeq 1t_i^c_i(sigma)right)=expleft(sum_igeq 1 fract_ix^iiright) =prod_igeq 1 left(1+fract_ix^ii+fract_i^2x^2i2i^2+cdotsright)$$
    From here we see that the exponential generating function of permutations with distinct cycle sizes can be obtained by removing all terms where any $t_i$ has exponent $geq 2$, and then setting all $t_i=1$. So we get
    $$prod_igeq 1left(1+fracx^iiright)$$
    From here the methods of A Hybrid of Darboux's Method and Singularity Analysis in Combinatorial Asymptotics by P. Flajolet, E. Fusy, X. Gourdon, D. Panario, N. Pouyanne show that the coefficient of $x^n$ is asymptotically equal to
    $$e^-gamma+frace^-gamman+Oleft(fraclog nn^2right)$$
    which means that our desired number of permutations is asymptotically given by $$n!left(e^-gamma+frace^-gamman+Oleft(fraclog nn^2right)right).$$
    Notice that in section 3 they actually provide much more refined asymptotics, in case you wanted more terms. Moreover, I believe their method should let you compute the asymptotics for permutations with cycles of the same length appearing at most $k$ times whose generating function is given by
    $$prod_igeq 1left(1+fracx^ii+fracx^2i2i^2+cdots +fracx^kik!i^kright)$$
    from the same considerations as above.






    share|cite|improve this answer
























      up vote
      4
      down vote













      If we denote by $c_i(sigma)$ the number of cycles of length $i$ in $sigma$, we can write the exponential generating function of permutations with cycle statistics as
      $$sum_ngeq 1sum_sigmain S_nleft(fracx^nn!prod_igeq 1t_i^c_i(sigma)right)=expleft(sum_igeq 1 fract_ix^iiright) =prod_igeq 1 left(1+fract_ix^ii+fract_i^2x^2i2i^2+cdotsright)$$
      From here we see that the exponential generating function of permutations with distinct cycle sizes can be obtained by removing all terms where any $t_i$ has exponent $geq 2$, and then setting all $t_i=1$. So we get
      $$prod_igeq 1left(1+fracx^iiright)$$
      From here the methods of A Hybrid of Darboux's Method and Singularity Analysis in Combinatorial Asymptotics by P. Flajolet, E. Fusy, X. Gourdon, D. Panario, N. Pouyanne show that the coefficient of $x^n$ is asymptotically equal to
      $$e^-gamma+frace^-gamman+Oleft(fraclog nn^2right)$$
      which means that our desired number of permutations is asymptotically given by $$n!left(e^-gamma+frace^-gamman+Oleft(fraclog nn^2right)right).$$
      Notice that in section 3 they actually provide much more refined asymptotics, in case you wanted more terms. Moreover, I believe their method should let you compute the asymptotics for permutations with cycles of the same length appearing at most $k$ times whose generating function is given by
      $$prod_igeq 1left(1+fracx^ii+fracx^2i2i^2+cdots +fracx^kik!i^kright)$$
      from the same considerations as above.






      share|cite|improve this answer






















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        If we denote by $c_i(sigma)$ the number of cycles of length $i$ in $sigma$, we can write the exponential generating function of permutations with cycle statistics as
        $$sum_ngeq 1sum_sigmain S_nleft(fracx^nn!prod_igeq 1t_i^c_i(sigma)right)=expleft(sum_igeq 1 fract_ix^iiright) =prod_igeq 1 left(1+fract_ix^ii+fract_i^2x^2i2i^2+cdotsright)$$
        From here we see that the exponential generating function of permutations with distinct cycle sizes can be obtained by removing all terms where any $t_i$ has exponent $geq 2$, and then setting all $t_i=1$. So we get
        $$prod_igeq 1left(1+fracx^iiright)$$
        From here the methods of A Hybrid of Darboux's Method and Singularity Analysis in Combinatorial Asymptotics by P. Flajolet, E. Fusy, X. Gourdon, D. Panario, N. Pouyanne show that the coefficient of $x^n$ is asymptotically equal to
        $$e^-gamma+frace^-gamman+Oleft(fraclog nn^2right)$$
        which means that our desired number of permutations is asymptotically given by $$n!left(e^-gamma+frace^-gamman+Oleft(fraclog nn^2right)right).$$
        Notice that in section 3 they actually provide much more refined asymptotics, in case you wanted more terms. Moreover, I believe their method should let you compute the asymptotics for permutations with cycles of the same length appearing at most $k$ times whose generating function is given by
        $$prod_igeq 1left(1+fracx^ii+fracx^2i2i^2+cdots +fracx^kik!i^kright)$$
        from the same considerations as above.






        share|cite|improve this answer












        If we denote by $c_i(sigma)$ the number of cycles of length $i$ in $sigma$, we can write the exponential generating function of permutations with cycle statistics as
        $$sum_ngeq 1sum_sigmain S_nleft(fracx^nn!prod_igeq 1t_i^c_i(sigma)right)=expleft(sum_igeq 1 fract_ix^iiright) =prod_igeq 1 left(1+fract_ix^ii+fract_i^2x^2i2i^2+cdotsright)$$
        From here we see that the exponential generating function of permutations with distinct cycle sizes can be obtained by removing all terms where any $t_i$ has exponent $geq 2$, and then setting all $t_i=1$. So we get
        $$prod_igeq 1left(1+fracx^iiright)$$
        From here the methods of A Hybrid of Darboux's Method and Singularity Analysis in Combinatorial Asymptotics by P. Flajolet, E. Fusy, X. Gourdon, D. Panario, N. Pouyanne show that the coefficient of $x^n$ is asymptotically equal to
        $$e^-gamma+frace^-gamman+Oleft(fraclog nn^2right)$$
        which means that our desired number of permutations is asymptotically given by $$n!left(e^-gamma+frace^-gamman+Oleft(fraclog nn^2right)right).$$
        Notice that in section 3 they actually provide much more refined asymptotics, in case you wanted more terms. Moreover, I believe their method should let you compute the asymptotics for permutations with cycles of the same length appearing at most $k$ times whose generating function is given by
        $$prod_igeq 1left(1+fracx^ii+fracx^2i2i^2+cdots +fracx^kik!i^kright)$$
        from the same considerations as above.







        share|cite|improve this answer












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        answered 2 hours ago









        Gjergji Zaimi

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