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A162005
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The EG1 triangle.
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18
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1, 2, 1, 16, 28, 1, 272, 1032, 270, 1, 7936, 52736, 36096, 2456, 1, 353792, 3646208, 4766048, 1035088, 22138, 1, 22368256, 330545664, 704357760, 319830400, 27426960, 199284, 1, 1903757312, 38188155904, 120536980224, 93989648000
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OFFSET
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1,2
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COMMENTS
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We define the EG1 matrix by EG1[2m-1,1] = 2*eta(2m-1) and the recurrence relation EG1[2m-1,n] = EG1[2m-1,n-1] - EG1[2m-3,n-1]/(n-1)^2 for m = -2, -1, 0, 1, 2, .. and n = 2, 3, .., with eta(m) = (1-2^(1-m))*zeta(m) with eta(m) the Dirichlet eta function and zeta(m) the Riemann zeta function. For the EG2[2m,n] coefficients see A008955.
The n-th term of the row coefficients EG1[1-2*m,n] for m = 1, 2, .., can be generated with REG1(1-2*m,n) = (-1)^(m+1)*2^(1-m)*ECGP(1-2*m, n)*(1/n)*4^(-n)*(2*n)!/((n-1)!)^2 . For information about the ECGP polynomials see A094665 and the examples below.
We define the o.g.f.s. of the REG1(1-2*m,n) by GFREG1(z,1-2*m) = sum(REG1(1-2*m,n)* z^(n-1), n=1..infinity) for m = 1, 2, .., with GFREG1(z,1-2*m) = (-1)^(m+1)* RG(z,1-2*m)/ (2^(2*m-1)*(1-z)^((2*m+1)/2)). The RG(z,1-2m) polynomials led to the EG1 triangle.
We used the coefficients of the A156919 and A094665 triangles to determine those of the EG1 triangle, see the Maple program. The A156919 triangle gives information about the sums SF(p) = sum(n^(p-1)*4^(-n)*z^(n-1)*(2*n)!/((n-1)!)^2, n=1..infinity) for p= 0, 1, 2, .. .
The EG1 matrix is related to the ED2 array A167560 because sum(EG1(2*m-1,n)*z^(2*m-1), m=1..infinity) = ((2*n-1)!/(4^(n-1)*(n-1)!^2))*int(sinh(y*(2*z))/cosh(y)^(2*n),y=0..infinity).
(End)
Appears to equal triangle A322230 with rows read in reverse order. Triangle A322230 describes the e.g.f. S(x,k) = Integral C(x,k)*D(x,k)^2 dx, such that C(x,k)^2 - S(x,k)^2 = 1, and D(x,k)^2 - k^2*S(x,k)^2 = 1. - Paul D. Hanna, Dec 22 2018
Appears to equal triangle A325220, which has e.g.f. S(x,k) = -i * sn( i * Integral C(x,k) dx, k) such that C(x,k) = cn( i * Integral C(x,k) dx, k), where sn(x,k) and cn(x,k) are Jacobi Elliptic functions. - Paul D. Hanna, Apr 13 2019
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LINKS
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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.
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FORMULA
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A different form of the recurrence relation is EG1[1-2*m,n] = (EG1[3-2*m,n]-EG1[3-2*m,n+1])* (n^2) for m = 2, 3, .., with EG1[ -1,n] = (1/n)*4^(-n)*((2*n)!/(n-1)!^2).
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EXAMPLE
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The first few rows of the EG1 triangle are :
[1]
[2, 1]
[16, 28, 1]
[272, 1032, 270, 1]
The first few RG(z,1-2*m) polynomials are:
RG(z,-1) = 1
RG(z,-3) = 2+z
RG(z,-5) = 16+28*z+z^2
RG(z,-7) = 272+1032*z+270*z^2+z^3
The first few GFREG1(z,1-2*m) are:
GFREG1(z,-1) = (1)*(1)/(2*(1-z)^(3/2))
GFREG1(z,-3) = (-1)*(2+z)/(2^3*(1-z)^(5/2))
GFREG1(z,-5) = (1)*(16+28*z+z^2)/( 2^5*(1-z)^(7/2))
GFREG1(z,-7) = (-1)*(272+1032*z+270*z^2+z^3)/(2^7*(1-z)^(9/2))
The first few REG1(1-2*m,n) are:
REG1(-1,n) = (1/1)*(1)*(1/n)*4^(-n)*(2*n)!/(n-1)!^2
REG1(-3,n) = (-1/2)*(n) *(1/n)*4^(-n)*(2*n)!/(n-1)!^2
REG1(-5,n) = (1/4) *(n+3*n^2) *(1/n)*4^(-n)*(2*n)!/(n-1)!^2
REG1(-7,n) = (-1/8)*(4*n+15*n^2+15*n^3) *(1/n)*4^(-n)*(2*n)!/(n-1)!^2
The first few ECGP(1-2*m,n) polynomials are:
ECGP(-1,n) = 1
ECGP(-3,n) = n
ECGP(-5,n) = n+3*n^2
ECGP(-7,n) = 4*n+15*n^2+15*n^3
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MAPLE
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nmax:=7; mmax := nmax: imax := nmax: T1(0, x) := 1: T1(0, x+1) := 1: for i from 1 to imax do T1(i, x) := expand((2*x+1) * (x+1)*T1(i-1, x+1)-2*x^2*T1(i-1, x)): dx := degree(T1(i, x)): for k from 0 to dx do c(k) := coeff(T1(i, x), x, k) od: T1(i, x+1) := sum(c(j1)*(x+1)^(j1), j1=0..dx): od: for i from 0 to imax do for j from 0 to i do A083061(i, j) := coeff(T1(i, x), x, j) od: od: for n from 0 to nmax do for k from 0 to n do A094665(n+1, k+1) := A083061(n, k) od: od: A094665(0, 0) := 1: for n from 1 to nmax do A094665(n, 0) := 0 od: for m from 1 to mmax do A156919(0, m) := 0 end do: for n from 0 to nmax do A156919(n, 0) := 2^n end do: for n from 1 to nmax do for m from 1 to mmax do A156919(n, m) := (2*m+2)*A156919(n-1, m) + (2*n-2*m+1)*A156919(n-1, m-1) end do end do: for n from 0 to nmax do SF(n) := sum(A156919(n, k1)*z^k1, k1=0..n)/(2^(n+1)*(1-z)^((2*n+3)/2)) od: GFREG1(z, -1) := A156919(0, 0)*A094665 (0, 0) / (2*(1-z)^(3/2)): for m from 2 to nmax do GFREG1(z, 1-2*m) := simplify((-1)^(m+1)*2^(1-m)* sum(A094665(m-1, k2)*SF(k2), k2=1..m-1)) od: for m from 1 to mmax do g(m) := sort((numer ((-1)^(m+1)* GFREG1(z, 1-2*m))), ascending) od: for n from 1 to nmax do for m from 1 to n do a(n, m) := abs(coeff(g(n), z, m-1)) od: od: seq(seq(a(n, m), m=1..n), n=1..nmax);
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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