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Triangle where T(n,k) = n!*[x^k] ( x/(2*x + log(1-x)) )^(n+1), for n>=k>=0, read by rows.
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%I #3 Mar 30 2012 18:36:57

%S 1,1,1,2,3,5,6,12,23,41,24,60,130,255,469,120,360,870,1860,3679,6889,

%T 720,2520,6720,15540,32858,65247,123605,5040,20160,58800,146160,

%U 328734,689388,1371887,2620169,40320,181440,574560,1527120,3638376,8029980

%N Triangle where T(n,k) = n!*[x^k] ( x/(2*x + log(1-x)) )^(n+1), for n>=k>=0, read by rows.

%C Row sums are A112487. Main diagonal is A032188(n) = number of labeled series-reduced mobiles (circular rooted trees) with n leaves.

%F Main diagonal has e.g.f.: series_reversion[2*x+log(1-x)].

%e Triangle begins:

%e 1;

%e 1, 1;

%e 2, 3, 5;

%e 6, 12, 23, 41;

%e 24, 60, 130, 255, 469;

%e 120, 360, 870, 1860, 3679, 6889;

%e 720, 2520, 6720, 15540, 32858, 65247, 123605;

%e 5040, 20160, 58800, 146160, 328734, 689388, 1371887, 2620169; ...

%e Triangle is formed from powers of F(x) = x/(2*x + log(1-x)):

%e F(x)^1 = (1) + 1/2*x + 7/12*x^2 + 17/24*x^3 + 629/720*x^4 +...

%e F(x)^2 = (1 + x)/1! +17/12*x^2 + 2*x^3 + 671/240*x^4 ...

%e F(x)^3 = (2 + 3*x + 5*x^2)/2! + 4*x^3 + 1489/240*x^4 +...

%e F(x)^4 = (6 + 12*x + 23*x^2 + 41/6*x^3)/3! + 8351/720*x^4 +...

%e F(x)^5 = (24 + 60*x + 130*x^2 + 255*x^3 + 469*x^4)/4! +...

%o (PARI) {T(n,k)=local(x=X+X^2*O(X^(k+2)));n!*polcoeff((x/(2*x+log(1-x)))^(n+1),k,X)}

%Y Cf. A118788, A112487, A032188.

%K nonn,tabl

%O 0,4

%A _Paul D. Hanna_, Apr 29 2006