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A111492 Triangle read by rows: a(n,k) = (k-1)! * C(n,k).

%I A111492 #57 Jun 28 2025 16:47:27
%S A111492 1,2,1,3,3,2,4,6,8,6,5,10,20,30,24,6,15,40,90,144,120,7,21,70,210,504,
%T A111492 840,720,8,28,112,420,1344,3360,5760,5040,9,36,168,756,3024,10080,
%U A111492 25920,45360,40320,10,45,240,1260,6048,25200,86400,226800,403200,362880
%N A111492 Triangle read by rows: a(n,k) = (k-1)! * C(n,k).
%C A111492 For k > 1, a(n,k) = the number of permutations of the symmetric group S_n that are pure k-cycles.
%C A111492 Reverse signed array is A238363. For a relation to (Cauchy-Euler) derivatives of the Vandermonde determinant, see Chervov link. - _Tom Copeland_, Apr 10 2014
%C A111492 Dividing the k-th column of T by (k-1)! for each column generates A135278 (the f-vectors, or face-vectors for the n-simplices). Then ignoring the first column gives A104712, so T acting on the column vector (-0,d,-d^2/2!,d^3/3!,...) gives the Euler classes for hypersurfaces of degree d in CP^n. Cf. A104712 and Dugger link therein. - _Tom Copeland_, Apr 11 2014
%C A111492 With initial i,j,n=1, given the n X n Vandermonde matrix V_n(x_1,...,x_n) with elements a(i=row,j=column)=(x_j)^(i-1), its determinant |V_n|, and the column vector of n ones C=(1,1,...,1), the n-th row of the lower triangular matrix T is given by the column vector determined by (1/|V_n|) * V_n(:x_1*d/dx_1:,...,:x_n*d/dx_n:)|V_n| * C, where :x_j*d/dx_j:^n = (x_j)^n*(d/dx_j)^n. - _Tom Copeland_, May 20 2014
%C A111492 For some other combinatorial interpretations of the first three columns of T, see A208535 and the link to necklace polynomials therein. Because of the simple relation of the array to the Pascal triangle, it can easily be related to many other arrays, e.g., T(p,k)/(p*(k-1)!) with p prime gives the prime rows of A185158 and A051168 when the non-integers are rounded to 0. - _Tom Copeland_, Oct 23 2014
%H A111492 A. Chervov, <a href="http://mathoverflow.net/questions/78540/a-sum-involving-derivatives-of-vandermonde">A sum involving derivatives of Vandermonde</a>
%F A111492 a(n, k) = (k-1)!C(n, k) = P(n, k)/k.
%F A111492 E.g.f. (by columns) = exp(x)((x^k)/k).
%F A111492 a(n, 1) = A000027(n);
%F A111492 a(n, 2) = A000217(n-1);
%F A111492 a(n, 3) = A007290(n);
%F A111492 a(n, 4) = A033487(n-3).
%F A111492 a(n, n) = A000142(n-1);
%F A111492 a(n, n-1) = A001048(n-1) for n > 1.
%F A111492 Sum[a(n, k), {k, 1, n}] = A002104(n);
%F A111492 Sum[a(n, k), {k, 2, n}] = A006231(n).
%F A111492 a(n,k) = sum(j=k..n-1, j!/(j-k)!) (cf. Chervov link). - _Tom Copeland_, Apr 10 2014
%F A111492 From _Tom Copeland_, Apr 28 2014: (Start)
%F A111492 E.g.f. by row: [(1+t)^n-1]/t.
%F A111492 E.g.f. of row e.g.f.s: {exp[(1+t)*x]-exp(x)}/t.
%F A111492 O.g.f. of row e.g.f.s: {1/[1-(1+t)*x] - 1/(1-x)}/t.
%F A111492 E.g.f. of row o.g.f.s: -exp(x) * log(1-t*x). (End)
%e A111492 a(3,3) = 2 because (3-1)!C(3,3) = 2.
%e A111492 1;
%e A111492 2 1;
%e A111492 3 3 2;
%e A111492 4 6 8 6;
%e A111492 5 10 20 30 24;
%e A111492 6 15 40 90 144 120;
%e A111492 7 21 70 210 504 840 720;
%e A111492 8 28 112 420 1344 3360 5760 5040;
%e A111492 9 36 168 756 3024 10080 25920 45360 40320;
%t A111492 Flatten[Table[(k - 1)!Binomial[n, k], {n, 10}, {k, n}]]
%o A111492 (Magma) /* As triangle: */ [[Factorial(k-1)*Binomial(n,k): k in [1..n]]: n in [1.. 15]]; // _Vincenzo Librandi_, Oct 21 2014
%K A111492 nonn,tabl
%O A111492 1,2
%A A111492 _Ross La Haye_, Nov 15 2005