cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A178790 The arithmetic mean of (2*k+1)*A_k (k=0,...,n-1), where A_0,A_1,... are Apery numbers given by A005259.

Original entry on oeis.org

1, 8, 127, 2624, 61501, 1552760, 41186755, 1131614720, 31923047665, 919243356008, 26908963456783, 798379043762624, 23954974906866901, 725620080605773592, 22159617936375571627, 681528994326392115200, 21090805673899997148025, 656256696917886135153800
Offset: 1

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Author

Zhi-Wei Sun, Jun 14 2010

Keywords

Comments

Conjecture: the number a(n) = n^(-1)*Sum_{k=0..n-1}(2*k+1)*A_k is always an integer.
We can prove that for any prime p>3 we have a(p)=p (mod p^4).
Conjecture: If p=5,7 (mod 8) is a prime then sum_{k=0}^{p-1}A_k=0 (mod p^2); if p=1,3 (mod 8) is a prime greater than 3 and p=x^2+2y^2 with x,y integers then sum_{k=0}^{p-1}A_k=4x^2-2p (mod p^2).
a(n) is always an integer. The detailed proof can be found in the latest version of arXiv:1006.2776 . - Zhi-Wei Sun, Jun 17 2010

Examples

			For n=3 we have a(3)=(A_0+3A_1+5A_2)/3=(1+3*5+5*73)/3=127.
		

Crossrefs

Programs

  • GAP
    List([1..30], n-> Sum([0..n], k -> Binomial(n-1,k)*Binomial(n+k,k) *Binomial(n+k, 2*k+1)* Binomial(2*k,k) )); # G. C. Greubel, Jan 24 2019
  • Magma
    [(&+[Binomial(n-1,k)*Binomial(n+k,k)*Binomial(n+k, 2*k+1)* Binomial(2*k,k): k in [0..n-1]]): n in [1..30]]; // G. C. Greubel, Jan 24 2019
    
  • Maple
    G := (-1/2)*(3*x-3+(x^2-34*x+1)^(1/2))*(x+1)^(-2)*hypergeom([1/3,2/3],[1],(-1/2)*(x^2-7*x+1)*(x+1)^(-3)*(x^2-34*x+1)^(1/2)+(1/2)*(x^3+30*x^2-24*x+1)*(x+1)^(-3))^2;
    ogf := 2*x*G/(1-x)-Int((x+1)*G/(x-1)^2,x);
    series(ogf,x=0,25); # Mark van Hoeij, May 07 2013
  • Mathematica
    Apery[n_]:= Sum[Binomial[n+k,k]^2 Binomial[n,k]^2,{k,0,n}]; AA[n_]:= Sum[(2k+1)*Apery[k],{k,0,n-1}]/n; Table[AA[n],{n,1,25}]
    Table[Sum[(Binomial[n-1,k]*Binomial[n+k,k]*Binomial[n+k,2*k+1]* Binomial[2*k,k]), {k,0,n-1}], {n,1,30}] (* G. C. Greubel, Jan 24 2019 *)
  • PARI
    {a(n) = sum(k=0,n-1, binomial(n-1,k)*binomial(n+k,k)*binomial(n+k, 2*k+1)*binomial(2*k,k))}; \\ G. C. Greubel, Jan 24 2019
    
  • Sage
    [sum(binomial(n-1,k)*binomial(n+k,k)*binomial(n+k, 2*k+1)* binomial(2*k,k) for k in (0..n-1)) for n in (1..30)] # G. C. Greubel, Jan 24 2019
    

Formula

Recursion : (n+2)^3 *(n+3) *(2n+1) *a(n+3) = (n+2) *(2n+1) *(35*n^3+193*n^2+345*n+203) *a(n+2) -(n+1) *(2n+5) *(35*n^3+122*n^2+132*n+40) * a(n+1) +n *(n+1)^3 *(2n+5) *a(n).
a(n) = Sum(k=0..n-1, (binomial(n-1,k)* binomial(n+k,k)* binomial(n+k,2*k+1)*binomial(2*k,k)) ). - Zhi-Wei Sun, Jun 17 2010
a(n) = A189766(n) / n = trace( HilbertMatrix(n)^(-1) )/n. - Richard Penner, Jun 04 2011
a(n) = (1/n)*Sum_{k=0..n-1} (2*k+1)*binomial(n+k,2*k+1)^2*binomial(2*k, k)^2. - Richard Penner, Jun 04 2011
G.f.: 2*x*G/(1-x)-Int((x+1)*G/(x-1)^2,x) where G is the generating function of A005259. - Mark van Hoeij, May 07 2013
a(n) ~ 2^(1/4) * (1 + sqrt(2))^(4*n) / (16*(Pi*n)^(3/2)). - Vaclav Kotesovec, Jan 24 2019

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

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Author

T. D. Noe, May 02 2011

Keywords

Comments

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

Examples

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

Crossrefs

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

Programs

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

Formula

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.

A348419 Triangular table read by rows: T(n,k) is the k-th entry of the main diagonal of the inverse Hilbert matrix of order n.

Original entry on oeis.org

1, 4, 12, 9, 192, 180, 16, 1200, 6480, 2800, 25, 4800, 79380, 179200, 44100, 36, 14700, 564480, 3628800, 4410000, 698544, 49, 37632, 2857680, 40320000, 133402500, 100590336, 11099088, 64, 84672, 11430720, 304920000, 2134440000, 4249941696, 2175421248, 176679360
Offset: 1

Views

Author

Jianing Song, Oct 18 2021

Keywords

Examples

			The inverse Hilbert matrix of order 4 is given by
  [  16  -120   240  -140]
  [-120  1200 -2700  1680]
  [ 240 -2700  6480 -4200]
  [-140  1680 -4200  2800].
Hence the 4th row is 16, 1200, 6480, 2800.
The first 8 rows of the table are:
  1,
  4, 12,
  9, 192, 180,
  16, 1200, 6480, 2800,
  25, 4800, 79380, 179200, 44100,
  36, 14700, 564480, 3628800, 4410000, 698544,
  49, 37632, 2857680, 40320000, 133402500, 100590336, 11099088,
  64, 84672, 11430720, 304920000, 2134440000, 4249941696, 2175421248, 176679360,
  ...
		

Crossrefs

Cf. A189766 (row sums), A189765, A005249.
A210356 gives the maximum value of each row and A210357 gives the positions of the maximum values.
Main diagonal gives A000515(n-1).

Programs

  • Maple
    T:= n-> (M-> seq(M[i, i], i=1..n))(1/LinearAlgebra[HilbertMatrix](n)):
    seq(T(n), n=1..8);  # Alois P. Heinz, Jun 19 2022
  • Mathematica
    T[n_, k_] := Inverse[HilbertMatrix[n]][[k, k]]; Table[T[n, k], {n, 1, 8}, {k, 1, n}] // Flatten (* Amiram Eldar, Oct 18 2021 *)
  • PARI
    T(n,k) = (1/mathilbert(n))[k,k]
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