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%I #24 Apr 17 2023 16:30:14
%S 1,1,6,16,106,426,3076,15856,123516,757756,6315976,44203776,391582456,
%T 3043809016,28496668656,241563299776,2378813448976,21703877431056,
%U 223903020594016,2177251989389056,23448038945820576,241173237884726176
%N E.g.f.: exp(x+5/2*x^2).
%C Term-by-term square of sequence with e.g.f.: exp(x+m/2*x^2) is given by e.g.f.: exp(x/(1-m*x))/sqrt(1-m^2*x^2) for all m.
%C a(n) is also the number of square roots of any permutation in S_{5n} whose disjoint cycle decomposition consists of n cycles of length 5. - _Luis Manuel Rivera Martínez_, Feb 26 2015
%H Vincenzo Librandi, <a href="/A115331/b115331.txt">Table of n, a(n) for n = 0..200</a>
%H Jesús Leaños, Rutilo Moreno, and Luis Manuel Rivera-Martínez, <a href="http://arxiv.org/abs/1005.1531">On the number of mth roots of permutations</a>, arXiv:1005.1531 [math.CO], 2010-2011.
%H Jesús Leaños, Rutilo Moreno, and Luis Manuel Rivera-Martínez, <a href="http://ajc.maths.uq.edu.au/pdf/52/ajc_v52_p041.pdf"> On the number of mth roots of permutations </a>, Australas. J. Combin. 52 (2012), 41-54 (Theorems 1 and 2).
%F Term-by-term square equals A115332 which has e.g.f.: exp(x/(1-5*x))/sqrt(1-25*x^2).
%F a(n) ~ exp(sqrt(n/5)-n/2-1/20)*5^(n/2)*n^(n/2)/sqrt(2). - _Vaclav Kotesovec_, Oct 19 2012
%F a(n) = n!*Sum_{k=0..floor(n/2)}5^k/(2^k*k!*(n-2*k)!). - _Luis Manuel Rivera Martínez_, Feb 26 2015
%F O.g.f.: 1/(1-x - 5*x^2/(1-x - 10*x^2/(1-x - 15*x^2/(1-x - 20*x^2/(1-x - 25*x^2/(1-x -...)))))), a continued fraction (after _Paul Barry_ in A115327). - _Paul D. Hanna_, Mar 08 2015
%t Range[0, 20]! CoefficientList[Series[Exp[(x + 5 / 2 x^2)], {x, 0, 20}], x] (* _Vincenzo Librandi_, May 22 2013 *)
%o (PARI) a(n)=local(m=5);n!*polcoeff(exp(x+m/2*x^2+x*O(x^n)),n)
%Y Column k=5 of A359762.
%Y Cf. A115332.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jan 20 2006
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