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<item>
  <id>05937096</id>
  <dt>j</dt>
  <an>05937096</an>
  <augroup>
    <au>Stanley, Richard P.</au>
  </augroup>
  <ti>Two enumerative results on cycles of permutations.</ti>
  <so>Eur. J. Comb. 32, No. 6, 937-943 (2011).</so>
  <py>2011</py>
  <pu>Elsevier Science (Academic Press), London</pu>
  <lagroup>
    <la>EN</la>
  </lagroup>
  <ccgroup>
  </ccgroup>
  <utgroup>
  </utgroup>
  <cigroup>
    <ci>Zbl 1127.05001</ci>
  </cigroup>
  <ligroup>
    <li>doi:10.1016/j.ejc.2011.01.011</li>
  </ligroup>
  <abgroup>
    <ab>Summary: Answering a question of {\it M. B\'ona} [A walk through combinatorics. An introduction to enumeration and graph theory. 2nd ed. Hackensack, NJ: World Scientific. xviii, 469 p. (2006; Zbl 1127.05001)], it is shown that for $n\ge 2$ the probability that 1 and 2 are in the same cycle of a product of two $n$-cycles on the set $\{1,2,\dots,n\}$ is $1/2$ if $n$ is odd and ${1\over 2}- {2\over(n-1)(n+2)}$ if $n$ is even. Another result concerns the polynomial $$P_\lambda(q)= \sum_w q^{\kappa((1,2,\dots, n)\cdot w)},$$ where $w$ ranges over all permutations in the symmetric group ${\germ S}_n$ of cycle type $\lambda$, $(1,2,\dots,n)$ denotes the $n$-cycle $1\to 2\to\cdots\to n\to 1$, and $\kappa(v)$ denotes the number of cycles of the permutation $v$. A formula is obtained for $P_\lambda(q)$ from which it is deduced that all zeros of $P_\lambda(q)$ have real part 0.</ab>
    <rv></rv>
  </abgroup>
</item>