A181145 - OEIS

%I #2 Mar 30 2012 18:37:22

%S 1,1,4,1,1,12,27,12,1,1,24,134,236,134,24,1,1,40,410,1540,2380,1540,

%T 410,40,1,1,60,975,6260,18386,26216,18386,6260,975,60,1,1,84,1981,

%U 19320,91441,227052,306495,227052,91441,19320,1981,84,1,1,112,3612,49672

%N G.f.: exp( Sum_{n>=1} [Sum_{k=0..2n} C(2n,k)^2*y^k]*x^n/n ) = Sum_{n>=0,k=0..2n} T(n,k)*x^n*y^k, as a triangle of coefficients T(n,k) read by rows.

%C Compare g.f. to that of the triangle A034870:

%C * exp( Sum_{n>=1} [Sum_{k=0..2n} C(2n,k)*y^k]*x^n/n )

%C which consists of the even numbered rows of Pascal's triangle.

%F Row sums form A066357 (with offset), the number of ordered trees on 2n nodes with every subtree at the root having an even number of edges.

%e G.f.: A(x,y) = 1 + (1+ 4*y+ y^2)*x + (1 + 12*y+ 27*y^2+ 12*y^3+ y^4)*x^2 + (1+ 24*y+ 134*y^2+ 236*y^3+ 134*y^4+ 24*y^5+ y^6)*x^3 +...

%e The logarithm of the g.f. begins:

%e log(A(x,y)) = (1 + 2^2*y + y^2)*x

%e + (1 + 4^2*y + 6^2*y^2 + 4^2*y^3 + y^4)*x^2/2

%e + (1 + 6^2*y + 15^2*y^2 + 20^2*y^3 + 15^2*y^4 + 6^2*y^5 + y^6)*x^3/3

%e + (1 + 8^2*y + 28^2*y^2 + 56^2*y^3 + 70^2*y^4 + 56^2*y^5 + 28^2*y^6 + 8^2*y^7 + y^8)*x^4/4 +...

%e Triangle begins:

%e 1;

%e 1, 4, 1;

%e 1, 12, 27, 12, 1;

%e 1, 24, 134, 236, 134, 24, 1;

%e 1, 40, 410, 1540, 2380, 1540, 410, 40, 1;

%e 1, 60, 975, 6260, 18386, 26216, 18386, 6260, 975, 60, 1;

%e 1, 84, 1981, 19320, 91441, 227052, 306495, 227052, 91441, 19320, 1981, 84, 1;

%e 1, 112, 3612, 49672, 344260, 1312080, 2883562, 3740572, 2883562, 1312080, 344260, 49672, 3612, 112, 1; ...

%o (PARI) {T(n,k)=polcoeff(polcoeff(exp(sum(m=1,n,sum(j=0,2*m,binomial(2*m,j)^2*y^j)*x^m/m)+O(x^(n+1))),n,x),k,y)}

%Y Cf. A066357 (row sums), A181146 (main diagonal).

%Y Cf. variants: A181143, A181144, A001263, A034870.

%K nonn,tabl

%O 0,3

%A _Paul D. Hanna_, Oct 16 2010

%E Comment and example corrected by _Paul D. Hanna_, Oct 16 2010