A160474 The Zeta triangle.
-1, 51, -10, -10594, 2961, -210, 356487, -115940, 12642, -420, -101141295, 35804857, -4751890, 254562, -4620, 48350824787, -18071509911, 2689347661, -180909586, 5471466, -60060
Offset: 2
Examples
The first few rows of the triangle ZETA(n,m) with n=2,3,... and m=1,2,... are [ -1], [51, -10], [ -10594, 2961, -210], [356487, -115940, 12642, -420]. The first few ZETA(z;n) polynomials are ZETA(z;n=2) = -1, ZETA(z;n=3) = 51-10*z^2, ZETA(z;n=4) = -10594 + 2961*z^2 - 210*z^4. The first few CFN1(z;n) polynomials are CFN1(z;n=2) = (z^2-1), CFN1(z;n=3) = (z^4 - 5*z^2 + 4), CFN1(z;n=4) = (z^6 - 14*z^4 + 49*z^2 - 36). The first few generating functions GH(z;n) are GH(z;n=2) = (6*(z^2-1)*GH(z;n=1) + (-1)) / 9, GH(z;n=3) = (60*(z^4-5*z^2+4)*GH(z;n=1) + (51-10*z^2)) / 450, GH(z;n=4) = (1260*(z^6-14*z^4+49*z^2-36)*GH(z;n=1) + (-10594+2961*z^2-210*z^4))/99225.
Links
- M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972, Chapter 23, pp. 811-812.
- Mohammad K. Azarian, Problem 1218, Pi Mu Epsilon Journal, Vol. 13, No. 2, Spring 2010, p. 116. Solution published in Vol. 13, No. 3, Fall 2010, pp. 183-185.
- Johannes W. Meijer, The zeros of the Eta, Zeta, Beta and Lambda polynomials, jpg and pdf, Mar 03 2013.
- Johannes W. Meijer and N.H.G. Baken, The Exponential Integral Distribution, Statistics and Probability Letters, Volume 5, No.3, April 1987. pp 209-211.
Crossrefs
A160475 equals the first left hand column.
A160476 equals the first right hand column and 6*h(n).
A160477 equals the rows sums.
A160478 equals the p(n) sequence.
A160479 equals the ZL(n) sequence.
A001620 is the Euler-Mascheroni constant gamma.
The CFN1(z, n) and the cfn1(n, k) lead to A008955.
Cf. A162446 (ZG1 matrix)
Programs
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Maple
nmax:=7; 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 Zc(n) := (Omega(n)*2^(2*n-1))*2/((2*n+1)*(n)) end do: c(1) := denom(Zc(1)): for n from 2 to nmax do c(n) := lcm(c(n-1)*(n)*(2*n+1)/2, denom(Zc(n))); p(n) := c(n-1) end do: y(1):=Zc(1): for n from 1 to nmax-1 do y(n+1) := Zc(n+1) - ((2*n+2)/(2*n+3))*y(n) end do: for n from 1 to nmax do b(n) := 4^(-n)*(2*n+1)*n*denom(Omega(n)) end do: for n from 1 to nmax-1 do c(n+1) := lcm(c(n)*(n+1)*(2*n+3)/2, b(n+1)) end do: for n from 1 to nmax do cm(n) := c(n)*(1/6)* 4^n/(2*n+1)! end do: for n from 1 to nmax-1 do ZL(n+2) := cm(n+1)/cm(n) end do: mmax := nmax: for n from 2 to nmax do ZETA(n, 1) := p(n)*y(n-1): ZETA(n, n) := 0 end do: for m from 2 to mmax do for n from m+1 to nmax do ZETA(n, m) := ZL(n)*(ZETA(n-1, m-1) - (n-1)^2* ZETA(n-1, m)) end do end do; seq(seq(ZETA(n,m), m=1..n-1), n=2..nmax); # End first program (program edited, Johannes W. Meijer, Sep 20 2012) nmax1 := 10; m := 1; ZS1row := 1-2*m; with(combinat): t1 := 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: mmax1 := nmax1: for m1 from 1 to mmax1 do M(m1-1) := 2^(2*m1-2)/((2*m1-1)!) end do: for m1 from 1 to mmax1 do ZS1[ -2*m1+1, 1] := 2*(-bernoulli(2*m1)/(2*m1)) od: for n from 2 to nmax1 do for m1 from 1 to mmax1-n+1 do ZS1[-2*m1+1, n] := M(n-1)*sum((-1)^(k1+1)*t1(n-1, k1-1) * ZS1[2*k1-2*n-2*m1+1, 1], k1 = 1..n) od: od: seq(ZS1[1-2*m, n], n = 1..nmax1-m+1); # End second program (program edited, Johannes W. Meijer, Sep 20 2012)
Formula
We discovered a remarkable relation between the Zeta triangle coefficients ZETA(n,m) = ZL(n)*(ZETA(n-1,m-1)-(n-1)^2*ZETA(n-1,m)) for n = 3, 4, ... and m = 2, 3, .... See A160475 for ZETA(n,m=1) and furthermore ZETA(n,n) = 0 for n = 2, 3, ....
The generating functions GH(z;n) of the coefficients in the matrix columns are defined by
GH(z;n) = sum(ZS1[2*m-1,n]*z^(2*m-2), m=1..infinity), with n = 1, 2, 3, .... This definition, and our choice of ZS1[1,1] = 2*gamma, leads to GH(z;n=1) = (-Psi(1-z)-Psi(1+z)) with Psi(z) the digamma-function. Furthermore we discovered that GH(z;n) = GH(z;n-1)*(2*z^2/((2*n-1)*(n-1))-(2*n-2)/(2*n-1))+2*ZS1[ -1,n-1]/((2*n-1)*(n-1)) for n = 2, 3 , ..., with ZS1[ -1,n] = 2^(2*n-1)*A002195(n)/A002196(n) for n = 1, 2, ....
We found the following general expression for the GH(z;n) polynomials, for n = 2, 3, ...:
GH(z;n) = (h(n)*CFN1(z;n)*GH(z;n=1) + ZETA(z;n))/p(n) with
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