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%I #32 Feb 24 2024 10:36:44
%S 1,1,1,1,4,2,19,4,1,99,18,3,611,99,9,1,4376,612,48,4,35621,4376,306,
%T 16,1,324965,35620,2190,100,5,3285269,324965,17810,730,25,1,36462924,
%U 3285270,162480,5940,180,6,440840359,36462924,1642635,54160,1485,36,1
%N Triangle read by rows: T(n,k) is the number of permutations of [n] having k adjacent transpositions (0 <= k <= floor(n/2)). An adjacent transposition is a cycle of the form (i, i+1).
%C Row n contains 1 + floor(n/2) entries.
%C Sum of entries in row n = n! (A000142).
%C T(n,0) = A177249(n).
%C Sum_{k>=0} k*a(n,k) = (n-1)! (n >= 2).
%H Seiichi Manyama, <a href="/A177248/b177248.txt">Rows n = 0..200, flattened</a>
%H R. A. Brualdi and E. Deutsch, <a href="http://arxiv.org/abs/1005.0781">Adjacent q-cycles in permutations</a>, arXiv:1005.0781 [math.CO], 2010.
%F T(n,k) = Sum_{j=0..floor(n/2)} (-1)^(k+j)*binomial(j,k)*(n-j)!/j!.
%F G.f. of column k: (1/k!) * Sum_{j>=k} j! * x^(j+k) / (1+x^2)^(j+1). - _Seiichi Manyama_, Feb 24 2024
%e T(5,2)=3 because we have (12)(34)(5), (12)(3)(45), and (1)(23)(45).
%e Triangle starts:
%e 1;
%e 1;
%e 1, 1;
%e 4, 2;
%e 19, 4, 1;
%e 99, 18, 3;
%e 611, 99, 9, 1;
%p T := proc (n, k) options operator, arrow: sum((-1)^(k+j)*binomial(j, k)*factorial(n-j)/factorial(j), j = 0 .. floor((1/2)*n)) end proc: for n from 0 to 12 do seq(T(n, k), k = 0 .. floor((1/2)*n)) end do; # yields sequence in triangular form
%t T[n_, k_] := Sum[(-1)^(k + j)*Binomial[j, k]*(n - j)!/j!, {j, 0, n/2}];
%t Table[T[n, k], {n, 0, 12}, {k, 0, n/2}] // Flatten (* _Jean-François Alcover_, Nov 20 2017 *)
%o (PARI) T(n, k) = sum(j=0, n\2, (-1)^(k+j)*binomial(j,k)*(n-j)!/j!);
%o tabf(nn) = for (n=0, nn, for (k=0, n\2, print1(T(n,k), ", ")); print); \\ _Michel Marcus_, Nov 21 2017
%Y Columns k=0..3 give A177249, A370524, A370426, A370529.
%Y Cf. A000166, A008290, A177250, A177251, A177252, A177253.
%K nonn,tabf
%O 0,5
%A _Emeric Deutsch_, May 07 2010
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