%I #47 Sep 03 2022 07:06:56
%S 1,1,6,1,30,90,1,126,1260,2520,1,510,13230,75600,113400,1,2046,126720,
%T 1580040,6237000,7484400,1,8190,1171170,28828800,227026800,681080400,
%U 681080400,1,32766,10663380,494053560,6972966000,39502663200,95351256000,81729648000,1,131070,96461910,8203431600,196556560200,1882311631200,8266953895200,16672848192000,12504636144000
%N Triangle read by rows: Joffe's central differences of zero, T(n,k), 1 <= k <= n.
%C T(n,k) gives the number of ordered set partitions of the set {1,2,...,2*n} into k even sized blocks. An example is given below. Cf. A019538 and A156289. - _Peter Bala_, Aug 20 2014
%D H. T. Davis, Tables of the Mathematical Functions. Vols. 1 and 2, 2nd ed., 1963, Vol. 3 (with V. J. Fisher), 1962; Principia Press of Trinity Univ., San Antonio, TX, Vol. 2, p. 283.
%D S. A. Joffe, Calculation of the first thirty-two Eulerian numbers from central differences of zero, Quart. J. Pure Appl. Math. 47 (1914), 103-126.
%D S. A. Joffe, Calculation of eighteen more, fifty in all, Eulerian numbers from central differences of zero, Quart. J. Pure Appl. Math. 48 (1917-1920), 193-271.
%H Muniru A Asiru, <a href="/A241171/b241171.txt">Rows n = 1..50 of triangle, flattened</a>
%F T(n,k) = 0 if k <= 0 or k > n, = 1 if k=1, otherwise T(n,k) = k*(2*k-1)*T(n-1,k-1) + k^2*T(n-1,k).
%F Related to Euler numbers A000364 by A000364(n) = (-1)^n*Sum_{k=1..n} (-1)^k*T(n,k). For example, A000364(3) = 61 = 90 - 30 + 1.
%F From _Peter Bala_, Aug 20 2014: (Start)
%F T(n,k) = 1/(2^(k-1))*Sum_{j = 1..k} (-1)^(k-j)*binomial(2*k,k-j)*j^(2*n).
%F T(n,k) = k!*A156289(n,k) = k!*(2*k-1)!!*A036969.
%F E.g.f.: A(t,z) := 1/( 1 - t*(cosh(z) - 1) ) = 1 + t*z^2/2! + (t + 6*t^2)*z^4/4! + (t + 30*t^2 + 90*t^3)*z^6/6! + ... satisfies the partial differential equation d^2/dz^2(A) = D(A), where D = t^2*(2*t + 1)*d^2/dt^2 + t*(5*t + 1)*d/dt + t.
%F Hence the row polynomials R(n,t) satisfy the differential equation R(n+1,t) = t^2*(2*t + 1)*R''(n,t) + t*(5*t + 1)*R'(n,t) + t*R(n,t) with R(0,t) = 1, where ' indicates differentiation w.r.t. t. This is equivalent to the above recurrence equation.
%F Recurrence for row polynomials: R(n,t) = t*( Sum_{k = 1..n} binomial(2*n,2*k)*R(n-k,t) ) with R(0,t) := 1.
%F Row sums equal A094088(n) for n >= 1.
%F A100872(n) = (1/2)*R(n,2). (End)
%e Triangle begins:
%e 1,
%e 1, 6,
%e 1, 30, 90,
%e 1, 126, 1260, 2520,
%e 1, 510, 13230, 75600, 113400,
%e 1, 2046, 126720, 1580040, 6237000, 7484400,
%e 1, 8190, 1171170, 28828800, 227026800, 681080400, 681080400,
%e 1, 32766, 10663380, 494053560, 6972966000, 39502663200, 95351256000, 81729648000,
%e ...
%e From _Peter Bala_, Aug 20 2014: (Start)
%e Row 2: [1,6]
%e k Ordered set partitions of {1,2,3,4} into k blocks Number
%e - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
%e 1 {1,2,3,4} 1
%e 2 {1,2}{3,4}, {3,4}{1,2}, {1,3}{2,4}, {2,4}{1,3}, 6
%e {1,4}{2,3}, {2,3}{1,4}
%e (End)
%p T := proc(n,k) option remember;
%p if k > n then 0
%p elif k=0 then k^n
%p elif k=1 then 1
%p else k*(2*k-1)*T(n-1,k-1)+k^2*T(n-1,k); fi;
%p end: # Minor edit to make it also work in the (0,0)-offset case. _Peter Luschny_, Sep 03 2022
%p for n from 1 to 12 do lprint([seq(T(n,k), k=1..n)]); od:
%t T[n_, k_] /; 1 <= k <= n := T[n, k] = k(2k-1) T[n-1, k-1] + k^2 T[n-1, k]; T[_, 1] = 1; T[_, _] = 0; Table[T[n, k], {n, 1, 9}, {k, 1, n}] (* _Jean-François Alcover_, Jul 03 2019 *)
%o (Sage)
%o @cached_function
%o def A241171(n, k):
%o if n == 0 and k == 0: return 1
%o if k < 0 or k > n: return 0
%o return (2*k^2 - k)*A241171(n - 1, k - 1) + k^2*A241171(n - 1, k)
%o for n in (1..6): print([A241171(n, k) for k in (1..n)]) # _Peter Luschny_, Sep 06 2017
%o (GAP) Flat(List([1..10],n->List([1..n],k->1/(2^(k-1))*Sum([1..k],j->(-1)^(k-j)*Binomial(2*k,k-j)*j^(2*n))))); # _Muniru A Asiru_, Feb 27 2019
%Y Case m=2 of the polynomials defined in A278073.
%Y Cf. A000680 (diagonal), A094088 (row sums), A000364 (alternating row sums), A281478 (central terms), A327022 (refinement).
%Y Diagonals give A002446, A213455, A241172, A002456.
%Y Cf. A019538, A036969, A156289.
%K nonn,tabl
%O 1,3
%A _N. J. A. Sloane_, Apr 22 2014