%I
%S 1,2,6,16,46,132,384,1120,3278,9612,28236,83072,244752,722048,2132704,
%T 6306304,18666190,55300732,163968612,486528288,1444571068,4291629384,
%U 12756459936,37934818112,112855778768,335867740704,999895548736
%N First column and main diagonal of triangle A092686, in which the convolution of each row with {1,2} produces a triangle that, when flattened, equals the flattened form of A092686.
%C Conjecture: Limit n>infinity a(n)^(1/n) = 3.  _Vaclav Kotesovec_, Jun 29 2015
%H Vaclav Kotesovec, <a href="/A092687/b092687.txt">Table of n, a(n) for n = 0..850</a>
%F G.f. satisfies: A(x) = A( x^2/(12x) )/(12x). Recurrence: a(n) = Sum_{k=0..floor(n/2)} C(nk,k)*2^(n2k)*a(k).  _Paul D. Hanna_, Jul 10 2006
%t m = 27; A[_] = 1; Do[A[x_] = A[x^2/(12x)]/(12x) + O[x]^m // Normal, {m}]; CoefficientList[A[x], x] (* _JeanFrançois Alcover_, Nov 03 2019 *)
%o (PARI) T(n,k)=if(n<0k>n,0, if(n==0&k==0,1, if(n==1&k<=1,2, if(k==n,T(n,0), 2*T(n1,k)+T(n1,k+1)))))
%o a(n)=T(n,0)
%o for(n=0,30,print1(a(n),", "))
%o (PARI) a(n)=local(A=1+x);for(i=0,n\2,A=subst(A,x,x^2/(12*x+x*O(x^n)))/(12*x));polcoeff(A,n) \\ _Paul D. Hanna_, Jul 10 2006
%o (PARI) /* Using Recurrence: */
%o a(n)=if(n==0, 1, sum(k=0, n\2, binomial(nk, k)*2^(n2*k)*a(k)))
%o for(n=0,30,print1(a(n),", ")) \\ _Paul D. Hanna_, Jul 10 2006
%Y Cf. A092683, A092686, A092688, A092689.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Mar 04 2004
