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%I #7 Sep 20 2012 13:56:13
%S -1,51,-10594,356487,-101141295,48350824787,-2405967772180,
%T 5296878246375849,-24680641353374049205,12431632076904547636178,
%U -34807634670487142385955264,5037797143580320963623681605
%N First left hand column of the Zeta triangle A160474
%p nmax:=13; with(combinat): cfn1 := proc(n, k): sum((-1)^j*stirling1(n+1, n+1-k+j) * stirling1(n+1, n+1-k-j), j=-k..k) end proc: Omega(0):=1: for n from 1 to nmax do Omega(n) := (sum((-1)^(k1+n+1)*(bernoulli(2*k1)/(2*k1))*cfn1(n-1, n-k1), k1=1..n))/(2*n-1)! end do: for n from 1 to nmax do d(n):= 2^(2*n-1)*Omega(n) end do: for n from 2 to nmax do Zc(n-1) := d(n-1)*2/((2*n-1)*(n-1)) end do: c(1) := denom(Zc(1)): for n from 1 to nmax-1 do c(n+1) := lcm(c(n)*(n+1)*(2*n+3)/2, denom(Zc(n+1))); p(n+1) := c(n) end do: y(1) := Zc(1): for n from 1 to nmax-2 do y(n+1) := Zc(n+1)-((2*n+2)/(2*n+3))*y(n) end do: for n from 2 to nmax do ZETA(n, 1) := p(n)*y(n-1) end do: seq(ZETA(n, 1), n=2..nmax);
%p # edited, _Johannes W. Meijer_, Sep 20 2012
%Y A160474 is the Zeta triangle.
%K easy,sign
%O 2,2
%A _Johannes W. Meijer_, May 24 2009
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