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%I #39 Apr 21 2023 15:11:45
%S -1,51,-10,-10594,2961,-210,356487,-115940,12642,-420,-101141295,
%T 35804857,-4751890,254562,-4620,48350824787,-18071509911,2689347661,
%U -180909586,5471466,-60060
%N The Zeta triangle.
%C The coefficients of the ZS1 matrix are defined by ZS1[2*m-1,n] = (2^(2*m-1))*int(y^(2*m-1)/(sinh(y))^(2*n), y=0..infinity)/factorial(2*m-1) for m = 1, 2, 3, .. and n = 1, 2, 3, .. under the condition that n <= (m-1).
%C This definition leads to ZS1[2*m-1,n=1] = 2*zeta(2*m-1), for m = 2, 3, .. , and the recurrence relation ZS1[2*m-1,n]:=(1/(2*n-1))*((2/(n-1))*ZS1[2*m-3,n-1]-(2*n-2)*ZS1[2*m-1,n-1]). As usual zeta(m) is the Riemann zeta function. These two formulas enable us to determine the values of the ZS[2*m-1,n] coefficients, with m all integers and n all positive integers, but not for all. If we choose, somewhat but not entirely arbitrarily, ZS1[1,n=1] = 2*gamma, with gamma the Euler-Mascheroni constant, we can determine them all.
%C The coefficients in the columns of the ZS1 matrix, for m = 1, 2, 3, .., and n = 2, 3, 4 .. , can be generated with the GH(z;n) polynomials for which we found the following general expression GH(z;n) = (h(n)*CFN1(z;n)*GH(z;n=1) + ZETA(z;n))/p(n).
%C The CFN1(z;n) polynomials depend on the central factorial numbers A008955.
%C The ZETA(z;n) are the Zeta polynomials which lead to the Zeta triangle.
%C The zero patterns of the Zeta polynomials resemble a UFO. These patterns resemble those of the Eta, Beta and Lambda polynomials, see A160464, A160480 and A160487.
%C The first Maple algorithm generates the coefficients of the Zeta triangle. The second Maple algorithm generates the ZS1[2*m-1,n] coefficients for m= 0, -1, -2, .. .
%C The M(n) sequence, see the second Maple algorithm, leads to Gould's sequence A001316 and a sequence that resembles the denominators in Taylor series for tan(x), i.e., A156769(n).
%C Some of our results are conjectures based on numerical evidence.
%H M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972, Chapter 23, pp. 811-812.
%H Mohammad K. Azarian, <a href="https://www.jstor.org/stable/24340810">Problem 1218</a>, Pi Mu Epsilon Journal, Vol. 13, No. 2, Spring 2010, p. 116. <a href="https://www.jstor.org/stable/24337914">Solution</a> published in Vol. 13, No. 3, Fall 2010, pp. 183-185.
%H Johannes W. Meijer, The zeros of the Eta, Zeta, Beta and Lambda polynomials, <a href="/A160474/a160474.jpg">jpg</a> and <a href="/A160474/a160474.pdf">pdf</a>, Mar 03 2013.
%H Johannes W. Meijer and N.H.G. Baken, <a href="http://dx.doi.org/10.1016/0167-7152(87)90041-1">The Exponential Integral Distribution</a>, Statistics and Probability Letters, Volume 5, No.3, April 1987. pp 209-211.
%F 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, ....
%F We observe that the ZL(n) = A160479(n) sequence also rules the Lambda triangle A160487.
%F The generating functions GH(z;n) of the coefficients in the matrix columns are defined by
%F 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, ....
%F We found the following general expression for the GH(z;n) polynomials, for n = 2, 3, ...:
%F GH(z;n) = (h(n)*CFN1(z;n)*GH(z;n=1) + ZETA(z;n))/p(n) with
%F h(n) = 6*A160476(n) and p(n) = A160478(n).
%e The first few rows of the triangle ZETA(n,m) with n=2,3,... and m=1,2,... are
%e [ -1],
%e [51, -10],
%e [ -10594, 2961, -210],
%e [356487, -115940, 12642, -420].
%e The first few ZETA(z;n) polynomials are
%e ZETA(z;n=2) = -1,
%e ZETA(z;n=3) = 51-10*z^2,
%e ZETA(z;n=4) = -10594 + 2961*z^2 - 210*z^4.
%e The first few CFN1(z;n) polynomials are
%e CFN1(z;n=2) = (z^2-1),
%e CFN1(z;n=3) = (z^4 - 5*z^2 + 4),
%e CFN1(z;n=4) = (z^6 - 14*z^4 + 49*z^2 - 36).
%e The first few generating functions GH(z;n) are
%e GH(z;n=2) = (6*(z^2-1)*GH(z;n=1) + (-1)) / 9,
%e GH(z;n=3) = (60*(z^4-5*z^2+4)*GH(z;n=1) + (51-10*z^2)) / 450,
%e 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.
%p 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);
%p # End first program (program edited, _Johannes W. Meijer_, Sep 20 2012)
%p 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);
%p # End second program (program edited, _Johannes W. Meijer_, Sep 20 2012)
%Y A160475 equals the first left hand column.
%Y A160476 equals the first right hand column and 6*h(n).
%Y A160477 equals the rows sums.
%Y A160478 equals the p(n) sequence.
%Y A160479 equals the ZL(n) sequence.
%Y A001620 is the Euler-Mascheroni constant gamma.
%Y The M(n-1) sequence equals A001316(n-1)/A156769(n) (n>=1).
%Y The ZS1[ -1, n] and the Omega(n) coefficients lead to A002195 and A002196.
%Y The CFN1(z, n) and the cfn1(n, k) lead to A008955.
%Y Cf. The Eta, Beta and Lambda triangles A160464, A160480 and A160487.
%Y Cf. A162446 (ZG1 matrix)
%K uned,easy,sign,tabl
%O 2,2
%A _Johannes W. Meijer_, May 24 2009
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