A005249 -id:A005249 - cp's OEIS frontend

A117664 Denominator of the sum of all elements in the n X n Hilbert matrix M(i,j) = 1/(i+j-1), where i,j = 1..n.

1, 3, 10, 105, 252, 2310, 25740, 9009, 136136, 11639628, 10581480, 223092870, 1029659400, 2868336900, 11090902680, 644658718275, 606737617200, 4011209802600, 140603459396400, 133573286426580, 5215718803323600

Offset: 1

Alexander Adamchuk, Apr 11 2006

nonn

Sum_{j=1..n} Sum_{i=1..n} 1/(i+j-1) = A117731(n) / A117664(n) = 2n * H'(2n) = 2n * A058313(2n) / A058312(2n), where H'(2n) is 2n-th alternating sign Harmonic Number. H'(2n) = H(2n) - H(n), where H(n) is n-th Harmonic Number. - Alexander Adamchuk, Apr 23 2006

			For n=2, the 2 X 2 Hilbert matrix is [1, 1/2; 1/2, 1/3], so a(2) = denominator(1 + 1/2 + 1/2 + 1/3) = denominator(7/3) = 3.
The n X n Hilbert matrix begins:
    1 1/2 1/3 1/4  1/5  1/6  1/7  1/8 ...
  1/2 1/3 1/4 1/5  1/6  1/7  1/8  1/9 ...
  1/3 1/4 1/5 1/6  1/7  1/8  1/9 1/10 ...
  1/4 1/5 1/6 1/7  1/8  1/9 1/10 1/11 ...
  1/5 1/6 1/7 1/8  1/9 1/10 1/11 1/12 ...
  1/6 1/7 1/8 1/9 1/10 1/11 1/12 1/13 ...
  ...

Eric Weisstein's World of Mathematics, Hilbert Matrix.
Eric Weisstein's World of Mathematics, Harmonic Number

Cf. A091342, A098118, A111876, A082687, A086881, A005249, A001008, A002805.

Numerator is A117731(n).

Table[Denominator[Sum[1/(i + j - 1), {i, n}, {j, n}]], {n, 30}]

a(n) = A111876(n-1)/n.
a(n) = denominator( Sum_{j=1..n} Sum_{i=1..n} 1/(i+j-1) ). Numerator is A117731(n). - Alexander Adamchuk, Apr 23 2006
a(n) = denominator( Sum_{k=1..n} (2*k)/(n+k) ). - Peter Bala, Oct 10 2021

A189765 Triangle in which row n has the n(n+1)/2 elements of the lower triangular part of the inverse of the n-th order Hilbert matrix.

Original entry on oeis.org

1, 4, -6, 12, 9, -36, 192, 30, -180, 180, 16, -120, 1200, 240, -2700, 6480, -140, 1680, -4200, 2800, 25, -300, 4800, 1050, -18900, 79380, -1400, 26880, -117600, 179200, 630, -12600, 56700, -88200, 44100, 36, -630, 14700, 3360, -88200, 564480, -7560, 211680

Offset: 1

T. D. Noe, May 02 2011

The n-th order Hilbert matrix has elements h(i,j) = 1/(i+j-1) for 1 <= i,j <=n. Only the lower triangular matrix is shown because the Hilbert matrix and its inverse are symmetric. The n-th row begins with n^2 and ends with A000515(n+1).
The sums of select rows of the inverse matrix are sequences A002457, A002736, A002738, A007531, and A054559.
The largest magnitude in the matrix is A210356(n). - T. D. Noe, Mar 28 2012
The sum of the elements of the n-th matrix is n^2. - T. D. Noe, Apr 02 2012

			Row 3 is 9, -36, 192, 30, -180, 180 which corresponds to the inverse
  9  -36   30
-36  192 -180
30 -180  180

T. D. Noe, Rows n = 1..25, flattened
Eric W. Weisstein, MathWorld: Hilbert Matrix

Cf. A002457, A002736, A002738, A005249 (determinant), A007531, A054559, A189766 (trace).

lowerTri[m_List] := Module[{n = Length[m]}, Flatten[Table[Take[m[[i]], i], {i, n}]]]; Flatten[Table[lowerTri[Inverse[HilbertMatrix[n]]], {n, 6}]]

a(n,i,j) = (-1)^(i+j) (i+j-1) binomial(n+i-1, n-j) binomial(n+j-1, n-i) binomial(i+j-2, i-1)^2 is the (i,j) element of the inverse of the n-th Hilbert matrix.

A060739 a(n) = (-1)^(n(n-1)/2) * Product_{k=0,...,n-1} (n+k-1)!/((k!)^2 * (n-1-k)!).

Original entry on oeis.org

1, 1, -2, -36, 7200, 17640000, -560105280000, -239102222768640000, 1408147589778024775680000, 116620600756651855983415296000000, -137839975629646325813680872620851200000000, -2352568589682795058651211199786427114330521600000000

Offset: 0

Noam Katz (noamkj(AT)hotmail.com), Apr 25 2001

Let A_n be the matrix of size n X n defined by: A_n[i,j] = 1/(binomial coefficient i+j-2 over i-1) = 1/C(i+j-2,i-1) where 1 <= i,j <= n. The diagonals of this matrix are the reciprocals of the entries in the Pascal triangle. Then a(n) = 1/det(A_n) = det((A_n)^(-1)).
From the formula for a(n) it follows that the determinant of (A_n)^(-1) is an integer. By inspecting the values of (A_n)^(-1) for small values of n it looks like (A_n)^(-1) is actually a matrix of integers but I do not have a proof of this fact.
Let M_n be the n X n matrix with M_n(i,j)=i/(i+j); then |a(n-1)|=1/det(M_n). - Benoit Cloitre, Apr 21 2002
Also related to the multinomial coefficients (i+j)!/i!/j! : abs(a(n))=(1/detQ_n-1) where Q_n is the n X n matrix q(i,j)=i!j!/(i+j)! - Benoit Cloitre, May 30 2002
From Alexander Adamchuk, Nov 14 2009: (Start)
Also a(n) = (-1)^(n(n-1)/2) * Product[ Binomial[2k,k]^2/2, {k,1,n-1} ].
It is simpler definition of a(n).
It follows from the observation that Sqrt[ Abs[ a(n+1)/a(n)/2 ] ] = {1, 3, 10, 35, 126, 462, ...} = C(2n+1, n+1) = A001700. (End)

			Here is the matrix A_4 for n=4: [1, 1, 1, 1; 1, 1/2, 1/3, 1/4; 1, 1/3, 1/6, 1/10; 1, 1/4, 1/10, 1/20]; a(4) = 7200 because det(A_4) = 1/7200

Harry J. Smith, Table of n, a(n) for n=0,...,43
T. M. Richardson, The Reciprocal Pascal Matrix, arXiv preprint arXiv:1405.6315 [math.CO], 2014.
Doron Zeilberger, Reverend Charles to the aid of Major Percy and Fields-Medalist Enrico, arXiv:1405.6315 [math.CO], 2014.
Doron Zeilberger, Reverend Charles to the aid of Major Percy and Fields-Medalist Enrico, Amer. Math. Monthly 103 (1996), 501-502.

Cf. A005249, A067689.

Cf. A001700. [Alexander Adamchuk, Nov 14 2009]

A060739 := n->(-1)^(n*(n-1)/2) * mul( (n+k-1)!/((k!)^2 * (n-1-k)!), k=0..n-1);

a[n_] := (-1)^(n (n - 1)/2)*Product[ Multinomial[k, k, n - 1 - k], {k, 0, n - 1}]; Table[a[n], {n, 0, 11}] (* Jean-François Alcover, Dec 08 2011, after first formula *)

for(n=1,15,print1(1/matdet(matrix(n,n,i,j,i/(j+i))),",")) \\ See Cloitre's comment

{ for (n=0, 43, if (n<2, a=1, a=(-1)^(n\2)/matdet(matrix(n-1, n-1, i, j, i/(j+i)))); write("b060739.txt", n, " ", a); ) } \\ Harry J. Smith, Jul 10 2009

def A060739(n): return (-1)^(n//2)*A163085(2*(n-1))/factorial(n-1) if n > 0 else 1
[A060739(i) for i in (0..11)] # Peter Luschny, Sep 18 2012

If Multinomial[a, b, c] denotes the multinomial coefficient (a+b+c)! / (a! * b! * c!) (which is an integer) then : a(n) = (-1)^(n(n-1)/2) * Product k=0, ..., n-1 Multinomial[k, k, n-1-k] = (-1)^(n(n-1)/2) * product k=0, ..., n-1 (n+k-1)!/((k!)^2 * (n-1-k)!)
a(n) = (-1)^(n(n-1)/2) * Product[ Binomial[2k,k]^2/2, {k,1,n-1} ]. [Alexander Adamchuk, Nov 14 2009]
|a(n)| = A163085(2*(n-1))/(n-1)! for n > 0. - Peter Luschny, Sep 18 2012
|a(n)| ~ A^3 * 2^(2*n^2 - 3*n + 5/12) * exp(n - 1/4) / (Pi^n * n^(n - 1/4)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, May 19 2020

A189766 Trace of the inverse of the n-th order Hilbert matrix.

Original entry on oeis.org

1, 16, 381, 10496, 307505, 9316560, 288307285, 9052917760, 287307428985, 9192433560080, 295998598024613, 9580548525151488, 311414673789269713, 10158681128480830288, 332394269045633574405, 10904463909222273843200, 358543696456299951516425

Offset: 1

T. D. Noe, May 02 2011

nonn

See the Mathematica program for a formula in terms of a hypergeometric function.

Vincenzo Librandi, Table of n, a(n) for n = 1..600

Cf. A005249 (determinant), A189765 (inverse Hilbert matrix).

Table[Trace[Inverse[HilbertMatrix[n]]], {n, 20}] (* or *)
Table[n^2 HypergeometricPFQ[{1/2, 1-n, 1-n, 1+n, 1+n}, {1, 1, 1, 3/2}, 1], {n, 20}]

a(n) = trace(1/mathilbert(n)) \\ Jianing Song, Oct 18 2021

From Richard Penner, Jun 05 2011: (Start)
a(n) = n * A178790(n) = Sum_{k=0..n-1} (2*k+1)*binomial(n+k, 2*k+1)^2 * binomial(2*k,k)^2.
a(n) = Sum_{k=1..n} A005408(k)*A005259(k-1) = Sum_{k=0..n-1} (2*k+1) * Sum_{j=0..k} binomial(k+j,j)^2 * binomial(k,j)^2. (End)
Recurrence: (n-1)^3*(2*n-5)*a(n) = (2*n-5)*(35*n^3 - 122*n^2 + 132*n - 40)*a(n-1) - (2*n-1)*(35*n^3 - 193*n^2 + 345*n - 203)*a(n-2) + (n-2)^3*(2*n-1)*a(n-3). - Vaclav Kotesovec, Aug 18 2013
a(n) ~ 2^(1/4)*(17+12*sqrt(2))^n/(16*Pi^(3/2)*sqrt(n)). - Vaclav Kotesovec, Aug 18 2013

A061065 For n <= 6, entry of maximal modulus in the inverse of the n-th Hilbert matrix. For n >= 3, this is the (n-1,n-1)-th entry.

Original entry on oeis.org

1, 12, 192, 6480, 179200, 4410000, 100590336, 2175421248, 45229916160, 912328045200, 17965673440000, 346945899203904, 6592659294154752, 123580568462478400, 2289795064260480000, 42003815644116000000

Offset: 1

Roger Cuculière, May 28 2001

Incorrect version of the largest element in the inverse of Hilbert's matrix. See A210356 for the correct version. See A210357 for the location of the maximal value. - T. D. Noe and Clark Kimberling, Mar 28 2012

Harry J. Smith, Table of n, a(n) for n = 1..100

Cf. A005249, A000515, A210356, A210357.

{ for (n=1, 100, if (n>2, a=((2*n-2)^2)*(2*n-3)*binomial(2*n-4, n-2)^2, if (n==1, a=1, a=12)); write("b061065.txt", n, " ", a) ) } \\ Harry J. Smith, Jul 17 2009

For n >= 3, a(n) = ((2n-2)^2)*(2n-3)*C(2n-4, n-2)^2. - David Wasserman, Jun 08 2002

More terms from David Wasserman, Jun 08 2002

A069640 Let M_n be the n X n matrix with M_n(i,j)=1/(i+j+1); then a(n)=1/det(M_n).

Original entry on oeis.org

3, 240, 378000, 10668672000, 5175372787200000, 42202225467872870400000, 5708700736339601341845504000000, 12701009683686045652926579789004800000000, 462068939479146913162956288390362787269836800000000

Offset: 1

Benoit Cloitre, Apr 21 2002

Wolfram Research, 1991 Mathematica Conference, Elementary Tutorial Notes, Section 1, Introduction to Mathematica, Paul Abbott, page 19.

Robert Israel, Table of n, a(n) for n = 1..40

Cf. A000108, A003046, A005249, A067689, A074962.

seq(1/LinearAlgebra:-Determinant(LinearAlgebra:-HilbertMatrix(n,n,-1)),n=1..10); # Robert Israel, Sep 26 2018

Hilbert[n_Integer] := Table[1/(i + j + 1), {i, n}, {j, n}]; Table[ 1 / Det[ Hilbert[n]], {n, 1, 8}] (* Robert G. Wilson v, Mar 13 2004 *)
Table[(2*n+1)!!*(n!*Product[(2*k)!/k!/(k+1)!,{k,0,n}])^2,{n,1,11}] (* Alexander Adamchuk, May 17 2006 *)
Table[2^(2*n^2+2*n-1/12) * Glaisher^3 * BarnesG[n+3/2]^2 *(n!)^2 *(2*n+1)!!/(E^(1/4)*Pi^(n+1/2)*BarnesG[n+3]^2), {n, 1, 11}] (* Vaclav Kotesovec, Mar 09 2014 *)

for(n=1,10,print1(1/matdet(matrix(n,n,i,j,1/(i+j+1))),","))

a(n) = (2*n+1)!!*(n!*Product[(2*k)!/k!/(k+1)!,{k,0,n}])^2. a(n) = (2*n+1)!!*(n!*A003046(n))^2, where A003046(n)is the Product of first n Catalan numbers A000108(n). a(n) = (2*n+1)!*n!/(2^n)*A003046(n)^2. - Alexander Adamchuk, May 17 2006

a(n) ~ A^3 * 2^(2*n^2+3*n+11/12) / (exp(1/4) * n^(7/4) * Pi^(n+1)), where A = 1.2824271291... is the Glaisher-Kinkelin constant (see A074962). - Vaclav Kotesovec, Nov 13 2014

A069651 For n >= 1, let M_n be the n X n matrix with M_n(i,j) = i^2/(i+j); then a(n) = 1/det(M_n). Also, a(0) = 1 by convention.

Original entry on oeis.org

1, 2, 18, 1200, 735000, 4667544000, 332086420512000, 279394363051195392000, 2892376010829659126572800000, 379850021025259936655866602240000000, 648304836222110631242066578424390188032000000

Offset: 0

Benoit Cloitre, Apr 21 2002

Also, determinant of the inverse of the (n+1)-st Hilbert matrix, divided by (2n+1)!. - Robert G. Wilson v, Feb 02 2004

Also, inverse of determinant of the matrix M_n(i,j) = i*j/(i+j). - Harry Richman, Aug 19 2019

Cf. A000142, A005249, A163085.

Table[1/((2n - 1)!Det[Table[1/(i + j - 1), {i, n}, {j, n}]]), {n, 10}] (* Robert G. Wilson v, Feb 02 2004 *)
Table[(n + 1)!/(2*n + 1)!*Product[Binomial[i, Floor[i/2]], {i, 1, 2*n + 1}], {n, 0, 10}] (* Stefan Steinerberger, Feb 26 2008 *)

for(n=1,15,print1(1/matdet(matrix(n,n,i,j,i^2/(j+i))),","))

def A069651(n): return A163085(2*n+1)/factorial(2*n+1)
[A069651(n) for n in (0..10)] # Peter Luschny, Sep 18 2012

a(n) = A005249(n)/A000142(n). - Robert G. Wilson v, Feb 02 2004
a(n) = (n+1)!/(2*n+1)! * Product[Binomial(i,Floor(i/2)), {i,1,2*n+1}]. - Stefan Steinerberger, Feb 26 2008
a(n) = A163085(2*n+1)/(2*n+1)! = A163085(2*n)/factorial(n)^2. - Peter Luschny, Sep 18 2012

Edited by N. J. A. Sloane, Feb 25 2008

A091342 Given (1) f(h,j,a) = ( [ ((a/gcd(a,h)) / gcd(j+1,(a/gcd(a,h)))) * (h(j+1)) ] - [ ((a/gcd(a,h)) / gcd(j+1,(a/gcd(a,h)))) * (ja) ] ) / a then let (2) a(h) = d(h,j) = lcm( f(h,j,1) ... f(h,j,h) ).

Original entry on oeis.org

1, 3, 10, 105, 252, 2310, 25740, 45045, 680680, 11639628, 10581480, 223092870, 1029659400, 2868336900, 77636318760, 4512611027925, 4247163320400, 4011209802600, 140603459396400, 133573286426580, 5215718803323600

Offset: 1

Scott C. Macfarlan (scottmacfarlan(AT)covance.com), Mar 01 2004

Solves the following arithmetic problem. Let (3) q#(i/j)=(q+h) be defined thus: Given q, an unknown fraction, q=qd/d, of integer value (initially) and h equal to any desired integer value and j equal to any desired integer value and i equal to j(q+h)+h.
Let '#' denote the repeated addition of the denominator j to the denominator of q and the repeated addition of the numerator i to the numerator of q, each addition recursively replacing the prior q fractions numerator and denominator respectively. This results in a series of fractions of mostly non-integer values.
After the first three iterations for any case would result in the following fractions: 1) (qd+i) / (d+j) 2) ((qd+i)+i) / ((d+j)+j) 3) (((qd+i)+i)+i) / (((d+j)+j)+j)
The question is, what is the smallest initial denominator, d (of q=qd/d) that in the course of the repeated additions will result in fractions of integer value, where every integer, from the initial (q+1)...(q+h) will be formed?
For example, let q = 4, h = 5, so (q+h)=9 and j = 1, so that for i we have i = 14. So in terms of q#(i/j)=(q+h) we have 4#(14/1)=9.
In this sample, since h=5 and j=1 we get from (2) above that d(5,1) = 252 as the solution. Applying this solution, we see then that the initial numerator of q, or q*d, becomes 4*252 = 1008 and the initial denominator is 252. Alternatively, in terms of q#(i/j)=(q+h) we have (1008/252)#(14/1)=9. We see that the repeated additions yield:
1) 1008 +14 and 252 +1 ~ 1022/253
2) 1022 +14 and 253 +1 ~ 1036/254
...
27) 1372 +14 and 277 +1 ~ 1386/279
28) 1386 +14 and 278 +1 ~ 1400/280 =5
...
63) 1876 +14 and 314 +1 ~ 1890/315 =6
...
108) 2506 +14 and 359 +1 ~ 2520/360 =7
...
168) 3346 +14 and 419 +1 ~ 3360/420 =8
...
252) 4522 +14 and 503 +1 ~ 4536/504 =9
Note that h=5 and j=1 were chosen so that d(5,1) = a(5) of the sequence. Also note that by the definition of q#(i/j)=(q+h) all answers are only a function of h and j.
Note also that q=qd/d can also be expressed as 0=0d/d and that any q#(i/j)=(q+h) can be expressed as 0#(i/j)=h [after adjustments are made to i]. For instance 4#(14/1)=9 is the equivalent of 0#(10/1)=5 in terms of d(h,j).
Interestingly: For any q#(i/j)=p and r#(s/j)=t then (q+r)#((i+s)/j)=(p+t). Also for any q#(i/j)=p then (qr)#((ir)/j)=(pr).
The sequence A025558 can be calculated from this formula when h = j, in otherwords using the sequence of d(1, 1)...d(n, n). i.e. a(7) = 735 = d(7,7) = lcm(49,21,35,7,21,7,1) a(8) = 2240 = d(8,8) = lcm(64,28,16,10,32,416,1)
Denominator of the sum of all elements of n X n Hilbert Matrix M[i,j] with alternate signs. M[i,j] = 1/(i+j-1)(i,j = 1..n). - Alexander Adamchuk, Apr 11 2006

			a(5) = lcm(9,4,7,3) = 252
a(7) = lcm(13,6,11,5,9,4,1) = 25740
a(10)= lcm(19,9,17,4,3,7,13,3,11,1) = 11639628
a(14)= lcm(27,13,25,6,23,11,3,5,19,9,17,4,15,1) = 2868336900
n=2: HilbertMatrix[n,n]
1 1/2
1/2 1/3
so a(2) = Denominator[(1 - 1/2 - 1/2 + 1/3)] = Denominator[1/3] = 3.
The n X n Hilbert matrix begins:
1 1/2 1/3 1/4 1/5 1/6 1/7 1/8 ...
1/2 1/3 1/4 1/5 1/6 1/7 1/8 1/9 ...
1/3 1/4 1/5 1/6 1/7 1/8 1/9 1/10 ...
1/4 1/5 1/6 1/7 1/8 1/9 1/10 1/11 ...
1/5 1/6 1/7 1/8 1/9 1/10 1/11 1/12 ...
1/6 1/7 1/8 1/9 1/10 1/11 1/12 1/13 ...

Eric Weisstein's World of Mathematics, Hilbert Matrix.
Eric Weisstein's World of Mathematics, Harmonic Number.

Cf. A025558.

Cf. A098118, A086881, A005249, A001008, A002805.

Denominator[Table[Sum[Sum[(-1)^(i+j)*1/(i+j-1),{i,1,n}],{j,1,n}],{n,1,40}]] (* Alexander Adamchuk, Apr 11 2006 *)

a(n) = Denominator[Sum[Sum[(-1)^(i+j)*1/(i+j-1),{i,1,n}],{j,1,n}]]. - Alexander Adamchuk, Apr 11 2006

A278838 a(n) = det M_n where M_n is the n X n matrix m(i,j) = A000041(i+j).

Original entry on oeis.org

1, 2, 1, -2, 2, 3, 0, -3, -1, 4, -3, -3, 2, -1, -12, 12, 11, 6, -5, 0, 5, -4, -9, -11, 1, 4, -20, -20, -4, 9, -18, -27, 8, 52, -73, 83, 245, 88, -60, -217, -157, 74, -30, -99, 57, 74, -29, -36, 101, 320, -205, -206, 125, -109, -27, 139, -203, -644, -629, 723

Offset: 0

Vaclav Kotesovec, Nov 29 2016

sign

Vaclav Kotesovec, Table of n, a(n) for n = 0..500

Cf. A000041, A005249, A071543, A204249, A278839, A278840.

Flatten[{1, Table[Det[Table[PartitionsP[i+j], {i, n}, {j, n}]], {n, 1, 100}]}]

A278839 a(n) = det M_n where M_n is the n X n matrix m(i,j) = A000009(i+j).

Original entry on oeis.org

1, 1, -2, -1, 1, -1, 0, 1, 2, 1, -1, -1, -1, 3, -3, -7, -2, 3, -1, 0, 1, 1, -2, 2, 3, -2, 0, 0, -2, -3, -1, 0, 9, -5, -4, 0, 1, -1, -3, 1, 4, 3, 3, -7, -3, 3, 5, -48, 75, 143, 194, -272, 62, -31, -65, 46, 22, 3, -10, 2, 15, -15, -13, -2, 11, -1, -35, -26, 108

Offset: 0

Vaclav Kotesovec, Nov 29 2016

sign

Vaclav Kotesovec, Table of n, a(n) for n = 0..500

Cf. A000009, A005249, A071543, A204249, A278838, A278841.

Flatten[{1, Table[Det[Table[PartitionsQ[i+j], {i, n}, {j, n}]], {n, 1, 100}]}]

cp's OEIS Frontend

A117664 Denominator of the sum of all elements in the n X n Hilbert matrix M(i,j) = 1/(i+j-1), where i,j = 1..n.

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A189765 Triangle in which row n has the n(n+1)/2 elements of the lower triangular part of the inverse of the n-th order Hilbert matrix.

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A060739 a(n) = (-1)^(n(n-1)/2) * Product_{k=0,...,n-1} (n+k-1)!/((k!)^2 * (n-1-k)!).

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A189766 Trace of the inverse of the n-th order Hilbert matrix.

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A061065 For n <= 6, entry of maximal modulus in the inverse of the n-th Hilbert matrix. For n >= 3, this is the (n-1,n-1)-th entry.

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A069640 Let M_n be the n X n matrix with M_n(i,j)=1/(i+j+1); then a(n)=1/det(M_n).

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Maple

Mathematica

PARI

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A069651 For n >= 1, let M_n be the n X n matrix with M_n(i,j) = i^2/(i+j); then a(n) = 1/det(M_n). Also, a(0) = 1 by convention.

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Mathematica