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%I #13 Dec 11 2020 03:32:37
%S 4,29,184,1148,7228,46224,300476,1983102,13266032,89795420,614058228,
%T 4236652416,29457698192,206215486597,1452248529432,10281676045348,
%U 73137772914324,522472109334560,3746685545297640,26961148855455180,194626321451800800,1409026233004925340
%N Number of partitions of the 2-dimensional hypercube resulting from a sequence of n bisections, each of which splits any part perpendicular to any of the axes, such that each axis is used at least once.
%H Alois P. Heinz, <a href="/A258416/b258416.txt">Table of n, a(n) for n = 2..1000</a>
%F From _Vaclav Kotesovec_, May 29 2015: (Start)
%F Recurrence: 5*(n-2)*(n-1)*n*(n+1)*(13616*n^4 - 138092*n^3 + 514558*n^2 - 835288*n + 498441)*a(n) = - 6*(n-2)*(n-1)*n*(27232*n^5 - 289800*n^4 + 1170888*n^3 - 2195854*n^2 + 1802270*n - 411881)*a(n-1) + 16*(n-2)*(n-1)*(544640*n^6 - 6612960*n^5 + 32102192*n^4 - 79406652*n^3 + 104891690*n^2 - 69498516*n + 17766135)*a(n-2) - 8*(n-2)*(2*n - 5)*(1524992*n^6 - 18516288*n^5 + 89869136*n^4 - 222469596*n^3 + 295082666*n^2 - 197989116*n + 52268391)*a(n-3) - 16*(2*n - 7)*(2*n - 5)*(4*n - 13)*(4*n - 11)*(13616*n^4 - 83628*n^3 + 181978*n^2 - 165984*n + 53235)*a(n-4).
%F a(n) ~ c * d^n / (sqrt(Pi) * n^(3/2)), where d = 7.721133226857077553917531558... is the root of the equation 256 + 512*d - 32*d^2 - 5*d^3 = 0, c = 1.11097484883257916279675191289... is the root of the equation -8 + 364*c^2 - 518*c^4 + 185*c^6 = 0.
%F (End)
%p b:= proc(n, k, t) option remember; `if`(t=0, 1, `if`(t=1,
%p A(n-1, k), add(A(j, k)*b(n-j-1, k, t-1), j=0..n-2)))
%p end:
%p A:= proc(n, k) option remember; `if`(n=0, 1,
%p -add(binomial(k, j)*(-1)^j*b(n+1, k, 2^j), j=1..k))
%p end:
%p T:= proc(n, k) option remember;
%p add(A(n, k-i)*(-1)^i*binomial(k, i), i=0..k)
%p end:
%p a:= n-> T(n, 2):
%p seq(a(n), n=2..25);
%t b[n_, k_, t_] := b[n, k, t] = If[t == 0, 1, If[t == 1, A[n - 1, k], Sum[A[j, k]*b[n - j - 1, k, t - 1], {j, 0, n - 2}]]];
%t A[n_, k_] := A[n, k] = If[n == 0, 1, -Sum[Binomial[k, j]*(-1)^j*b[n + 1, k, 2^j], {j, 1, k}]];
%t T[n_, k_] := Sum[A[n, k - i]*(-1)^i*Binomial[k, i], {i, 0, k}];
%t a[n_] := T[n, 2];
%t a /@ Range[2, 25] (* _Jean-François Alcover_, Dec 11 2020, after _Alois P. Heinz_ *)
%Y Column k=2 of A255982.
%K nonn
%O 2,1
%A _Alois P. Heinz_, May 29 2015
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