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A383235 Triangle read by rows: T(n,k) = 2*floor(k/2)*T(n-1,k) + T(n-1,k-1), 0 <= k <= n.

%I A383235 #20 May 06 2025 08:46:00
%S A383235 1,0,1,0,0,1,0,0,2,1,0,0,4,4,1,0,0,8,12,8,1,0,0,16,32,44,12,1,0,0,32,
%T A383235 80,208,92,18,1,0,0,64,192,912,576,200,24,1,0,0,128,448,3840,3216,
%U A383235 1776,344,32,1,0,0,256,1024,15808,16704,13872,3840,600,40,1
%N A383235 Triangle read by rows: T(n,k) = 2*floor(k/2)*T(n-1,k) + T(n-1,k-1), 0 <= k <= n.
%C A383235 Row sums give A007472.
%C A383235 The recurrence is analogous to that of Stirling numbers of the 2nd kind (A048993), but with a parity constraint. For even columns, T(n+1, 2*k) = 2*k*T(n, 2*k) + T(n, 2*k-1); for odd columns, T(n+1, 2*k+1) = 2*k*T(n, 2*k+1) + T(n, 2*k).
%C A383235 T(n,k) appear as coefficients in the following polynomials (which themselves are coefficients  of t^n in the bivariate e.g.f.:
%C A383235 P_0(x) = 1
%C A383235 P_1(x) = 0 + x
%C A383235 P_2(x) = 0 + 0 + x^2
%C A383235 P_3(x) = 0 + 0 + 2x^2 + x^3
%C A383235 P_4(x) = 0 + 0 + 4x^2 + 4x^3 + x^4
%C A383235 P_5(x) = 0 + 0 + 8x^2 + 12x^3 + 8x^4 + x^5
%C A383235 P_6(x) = 0 + 0 + 16x^2 + 32x^3 + 44x^4 + 12x^5 + x^6
%C A383235 P_n(1) gives A007472.
%C A383235 T(n,k) enumerates the number of ways to partition n labeled objects in k compartments of urns with two compartments each with the following constraints:
%C A383235 - you can initially place a marble in any compartment of any urn
%C A383235 - you cannot use the same compartment in an urn again until you've used its other compartment
%C A383235 - you can freely place objects in any compartment of urns where both compartments are already used
%C A383235 - you cannot open a new urn until both compartments of all previously used urns are filled
%C A383235 Examples:
%C A383235 For T(3,2) we have two ways to place the objects in two compartments of a single urn
%C A383235 - {13|2}
%C A383235 - {1|23}
%C A383235 For T(3,3) we have one way to place each object in its own compartment (2 urns, 3 compartments used):
%C A383235 - {1|2}{3|}
%C A383235 For 4 objects we have:
%C A383235 - 4 cases for Z(4,2):
%C A383235 {134|2}
%C A383235 {14|23}
%C A383235 {13|24}
%C A383235 {1|234}
%C A383235 - 4 cases for Z(4,3):
%C A383235 {{13|2},{4|}}
%C A383235 {{1|23},{4|}}
%C A383235 {{14|2},{3|}}
%C A383235 {{1|24},{3|}}
%C A383235 - 1 case for Z(4,4):
%C A383235 {{1|2}, {3|4}}
%C A383235 Then A007472 counts the total number of ways to partition n labeled objects in such nonempty two-compartment urns.
%H A383235 Ven Popov, <a href="https://www.wolframcloud.com/obj/vpopov0/Published/marray-recurrence-triangle.nb">m-arity Stirling like arrays and generalization of the Bell numbers</a>
%F A383235 T(n, k) = 2*Floor(k/2)*T(n-1, k) + T(n-1, k-1), n > 0; T(0, k) = 0, k > 0; T(0, 0) = 1.
%F A383235 E.g.f.: A[x_, t_] := x*(BesselK[0,x] + BesselK[1,x])*BesselI[0,x*Exp[t]] + x*(BesselI[1,x] - BesselI[0,x])*BesselK[0, x Exp[t]] where T(n,k) are the coefficients of x^k for t^n. The much simpler formula BesselI[0,x*Exp[t]] produces the same coefficients but with carrier alternating BesselI[0,x] and BesselI[1,x] functions at even and odd coefficients.
%e A383235 The triangle T(n,k) begins:
%e A383235   n\k 0 1    2     3      4       5       6      7      8     9   10 11 12
%e A383235   0:  1
%e A383235   1:  0 1
%e A383235   2:  0 0    1
%e A383235   3:  0 0    2     1
%e A383235   4:  0 0    4     4      1
%e A383235   5:  0 0    8    12      8       1
%e A383235   6:  0 0   16    32     44      12       1
%e A383235   7:  0 0   32    80    208      92      18      1
%e A383235   8:  0 0   64   192    912     576     200     24      1
%e A383235   9:  0 0  128   448   3840    3216    1776    344     32     1
%e A383235   10: 0 0  256  1024  15808   16704   13872   3840    600    40    1
%e A383235   11: 0 0  512  2304  64256   82624   99936  36912   8640   920   50  1
%e A383235   12: 0 0 1024  5120 259328  394752  682240 321408 106032 16000 1420 60  1
%e A383235 ------------------------------------------------------------------------
%e A383235 Recurrence:
%e A383235 S(5,3) = 2*Floor(3/2)*S(4,3) + S(4,2) = 2*4 + 4 = 12.
%e A383235 S(5,4) = 2*Floor(4/2)*S(4,4) + S(4,3) = 4*1 + 4 = 8.
%t A383235 Z[0,0,m_]=1;
%t A383235 Z[0,k_,m_]:=0/;k>0;
%t A383235 Z[n_,0,m_]:=0/;n>0;
%t A383235 Z[n_,k_,m_]:=Z[n,k,m]=m*Floor[k/m]*Z[n-1,k,m]+Z[n-1,k-1,m];
%t A383235 Flatten[Table[Z[n,k,2],{n,0,12},{k,0,n}]]
%Y A383235 Cf. A048993. Row sums give A007472. Column T(n+2,2) gives A000079(n). Column T(n+2,3) gives A001787(n). Column T(n+2,4) gives A100575(n). Column T(n+5,5)*4 gives A158681(n). T(n+1,n) gives A007590(n).
%K A383235 nonn,tabl
%O A383235 0,9
%A A383235 _Ven Popov_, Apr 20 2025