A160466 Row sums of the Eta triangle A160464.

-1, -9, -87, -2925, -75870, -2811375, -141027075, -18407924325, -1516052821500, -153801543183750, -18845978136851250, -2744283682352086875, -468435979952504313750, -92643070481933918821875

Offset: 2

Author

Johannes W. Meijer, May 24 2009

Keywords

• easy
• sign

Comments

It is conjectured that the row sums of the Eta triangle depend on five different sequences.

Two Maple algorithms are given. The first one gives the row sums according to the Eta triangle A160464 and the second one gives the row sums according to our conjecture.

Crossrefs

A160464 is the Eta triangle.

Row sum factors A119951, A000466, A043529, A045896 and A160467.

Programs

Maple

nmax:=15; c(2) := -1/3: for n from 3 to nmax do c(n):=(2*n-2)*c(n-1)/(2*n-1)-1/ ((n-1)*(2*n-1)) end do: for n from 2 to nmax do GCS(n-1) := ln(1/(2^(-(2*(n-1)-1-floor(ln(n-1)/ ln(2))))))/ln(2); p(n):=2^(-GCS(n-1))*(2*n-1)!; ETA(n, 1) := p(n)*c(n) end do: mmax:=nmax: for m from 2 to mmax do ETA(2, m) := 0 end do: for n from 3 to nmax do for m from 2 to mmax do q(n) := (1+(-1)^(n-3)*(floor(ln(n-1)/ln(2)) - floor(ln(n-2)/ln(2)))): ETA(n, m) := q(n)*(-ETA(n-1, m-1)+(n-1)^2*ETA(n-1, m)) end do end do: for n from 2 to nmax do s1(n):=0: for m from 1 to n-1 do s1(n) := s1(n) + ETA(n, m) end do end do: seq(s1(n), n=2..nmax);
# End first program.
nmax:=nmax; A160467 := proc(n): denom(4*(4^n-1)*bernoulli(2*n)/n) end: A043529 := proc(n): ceil(frac(log[2](n+1))+1) end proc: A000466 := proc(n): 4*n^2-1 end proc: A045896 := proc(n): denom((n)/((n+1)*(n+2))) end proc: A119951 := proc(n) : numer(sum(((2*k1)!/(k1!*(k1+1)!))/2^(2*(k1-1)), k1=1..n)) end proc: for n from 1 to nmax do SF(2*n+1):= A000466(n)/A043529(n-1); SF(2*n+2) := A045896(n-1)/A160467(n+1) end do: FF(2):=1: for n from 3 to nmax do FF(n) := SF(n) * FF(n-1) end do: for n from 2 to nmax do s2(n):= (-1)*A119951(n-1)*FF(n) end do: seq(s2(n), n=2..nmax);
# End second program.

Formula

Rowsums(n) = (-1) * A119951(n-1) * FF(n) for n >= 2.

FF(n) = SF(n) * FF(n-1) for n >= 3 with FF(2) =1.

SF(2*n) = A045896(n-2) / A160467(n) for n >= 2.

SF(2*n+1) = A000466(n) / A043529(n-1) for n >= 1.