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A326763 Number of permutations of length n and order at most 3 whose powers all avoid the pattern 132.

%I A326763 #20 Apr 14 2025 11:53:02
%S A326763 1,1,2,5,10,16,36,65,118,232,452,800,1622,3042,5758,11077,21712,40204,
%T A326763 79718,151628,292994,561954,1103786,2087696,4115506,7884446,15393710,
%U A326763 29592074,58229334,111422134,219575234,422888473,830617400,1602832900,3160618558,6092881976
%N A326763 Number of permutations of length n and order at most 3 whose powers all avoid the pattern 132.
%H A326763 Miklós Bóna and Rebecca Smith, <a href="https://doi.org/10.1016/j.disc.2019.06.029">Pattern Avoidance in Permutations and Their Squares</a>, Discrete Mathematics, 342 (2019), pp. 3194-3200; arXiv:<a href="https://arxiv.org/abs/1901.00026">1901.00026</a> [math.CO], 2019.
%H A326763 Amanda Burcroff and Colin Defant, <a href="https://doi.org/10.1016/j.disc.2020.112017">Pattern-Avoiding Permutation Powers</a>, Discrete Mathematics, 343 (2020), 112017; arXiv:<a href="https://arxiv.org/abs/1907.09451">1907.09451</a> [math.CO], 2019.
%F A326763 a(n) = A014495(n) + A370686(n), where the 1st (resp. 2nd) term counts 132-avoiding permutations of order 2 (resp. 1 or 3). - _Andrey Zabolotskiy_, Apr 13 2025
%o A326763 (SageMath)
%o A326763 def a(n):
%o A326763     return len([p for p in Permutations(n)
%o A326763         if p*p == Permutations(n).identity() and p.avoids([1, 3, 2])
%o A326763         or p*p*p == Permutations(n).identity() and p.avoids([1, 3, 2]) and (p*p).avoids([1, 3, 2])]) # _Andrey Zabolotskiy_, Apr 13 2025
%Y A326763 Cf. A326760, A326761, A326762, A001405, A014495, A370686.
%K A326763 nonn
%O A326763 0,3
%A A326763 _Amanda Burcroff_, Aug 15 2019
%E A326763 Terms a(16) onwards using the formula from _Andrey Zabolotskiy_, Apr 14 2025