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%I #5 Sep 18 2012 15:57:39
%S 12,1440,907200,101606400,100590336000,172613016576000,
%T 31415569016832000,256351043177349120000,4471274895099323351040000,
%U 8495422300688714366976000000,90272357367118278863486976000000
%N The p(n) sequence that is associated with the Lambda triangle A160487
%p nmax:=11; for n from 0 to nmax do cfn2(n, 0) := 1: cfn2(n, n) := (doublefactorial(2*n-1))^2 od: for n from 1 to nmax do for k from 1 to n-1 do cfn2(n, k) := (2*n-1)^2*cfn2(n-1, k-1) + cfn2(n-1, k) od: od: for n from 1 to nmax do Delta(n-1) := sum((1-2^(2*k1-1))* (-1)^(n+1)*(-bernoulli(2*k1)/(2*k1))*(-1)^(k1+n)*cfn2(n-1,n-k1), k1=1..n) /(2*4^(n-1)*(2*n-1)!); LAMBDA(-2, n) := sum(2*(1-2^(2*k1-1))*(-bernoulli(2*k1)/ (2*k1))*(-1)^(k1+n)* cfn2(n-1,n-k1), k1=1..n) / factorial(2*n-2) end do: Lcgz(2) := 1/12: f(2) := 1/12: for n from 3 to nmax do Lcgz(n) := LAMBDA(-2, n-1)/((2*n-2)*(2*n-3)): f(n) := Lcgz(n)-((2*n-3)/(2*n-2))*f(n-1) end do: for n from 1 to nmax do b(n) := denom(Lcgz(n+1)) end do: for n from 1 to nmax do b(n) := 2*n*denom(Delta(n-1))/2^(2*n) end do: p(2) := b(1): for n from 2 to nmax do p(n+1) := lcm(p(n)*(2*n)*(2*n-1), b(n)) end do: seq(p(n), n=2..nmax+1);
%Y A160487 is the Lambda triangle.
%Y Equals 6*(2*n-2)!*A160476(n).
%K easy,nonn
%O 2,1
%A _Johannes W. Meijer_, May 24 2009, Sep 18 2012
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