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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A143007 Square array, read by antidiagonals, where row n equals the crystal ball sequence for the 2*n-dimensional lattice A_n x A_n.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 13, 13, 1, 1, 25, 73, 25, 1, 1, 41, 253, 253, 41, 1, 1, 61, 661, 1445, 661, 61, 1, 1, 85, 1441, 5741, 5741, 1441, 85, 1, 1, 113, 2773, 17861, 33001, 17861, 2773, 113, 1, 1, 145, 4873, 46705, 142001, 142001, 46705, 4873, 145, 1
Offset: 0

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Author

Peter Bala, Jul 22 2008

Keywords

Comments

The A_n lattice consists of all vectors v = (x_1,...,x_(n+1)) in Z^(n+1) such that x_1 + ... + x_(n+1) = 0. The lattice is equipped with the norm ||v|| = 1/2*(|x_1| + ... + |x_(n+1)|). Pairs of lattice points (v,w) in the product lattice A_n x A_n have norm ||(v,w)|| = ||v|| + ||w||. Then the k-th term in the crystal ball sequence for the A_n x A_n lattice gives the number of such pairs (v,w) for which ||(v,w)|| is less than or equal to k.
This array has a remarkable relationship with Apery's constant zeta(3). The row (or column) and main diagonal entries of the array occur in series acceleration formulas for zeta(3). For row n entries there holds zeta(3) = (1+1/2^3+...+1/n^3) + Sum_{k >= 1} 1/(k^3*T(n,k-1)*T(n,k)). Also, as consequence of Apery's proof of the irrationality of zeta(3), we have a series acceleration formula along the main diagonal of the table: zeta(3) = 6 * sum {n >= 1} 1/(n^3*T(n-1,n-1)*T(n,n)). Apery's result appears to generalize to the other diagonals of the table. Calculation suggests the following result may hold: zeta(3) = 1 + 1/2^3 + ... + 1/k^3 + Sum_{n >= 1} (2*n+k)*(3*n^2 +3*n*k +k^2)/(n^3*(n+k)^3*T(n-1,n+k-1)*T(n,n+k)).
For the corresponding results for the constant zeta(2), related to the crystal ball sequences of the lattices A_n, see A108625. For corresponding results for log(2), coming from either the crystal ball sequences of the hypercubic lattices A_1 x ... x A_1 or the lattices of type C_n, see A008288 and A142992 respectively.

Examples

			The table begins
n\k|0...1.....2......3.......4.......5
======================================
0..|1...1.....1......1.......1.......1
1..|1...5....13.....25......41......61 A001844
2..|1..13....73....253.....661....1441 A143008
3..|1..25...253...1445....5741...17861 A143009
4..|1..41...661...5741...33001..142001 A143010
5..|1..61..1441..17861..142001..819005 A143011
........
Example row 1 [1,5,13,...]:
The lattice A_1 x A_1 is equivalent to the square lattice of all integer lattice points v = (x,y) in Z x Z equipped with the taxicab norm ||v|| = (|x| + |y|). There are 4 lattice points (marked with a 1 on the figure below) satisfying ||v|| = 1 and 8 lattice points (marked with a 2 on the figure) satisfying ||v|| = 2. Hence the crystal ball sequence for the A_1 x A_1 lattice begins 1, 1+4 = 5, 1+4+8 = 13, ... .
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . 2 . . . . .
. . . . 2 1 2 . . . .
. . . 2 1 0 1 2 . . .
. . . . 2 1 2 . . . .
. . . . . 2 . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
Row 1 = [1,5,13,...] is the sequence of partial sums of A008574; row 2 = [1,13,73,...] is the sequence of partial sums of A008530, so row 2 is the crystal ball sequence for the lattice A_2 x A_2 (the 4-dimensional di-isohexagonal orthogonal lattice).
Read as a triangle the array begins
n\k|0...1....2....3...4...5
===========================
0..|1
1..|1...1
2..|1...5....1
3..|1..13...13....1
4..|1..25...73...25...1
5..|1..41..253..253..41...1
		

Crossrefs

Cf. A001844 (row 1), A005259 (main diagonal), A008288, A008530 (first differences of row 2), A008574 (first differences of row 1), A085478, A108625, A142992, A143003, A143004, A143005, A143006, A143008 (row 2), A143009 (row 3), A142010 (row 4), A143011 (row 5).
Cf. A227845 (antidiagonal sums), A246464.
The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)

Programs

  • Magma
    A:= func< n,k | (&+[(Binomial(n,j)*Binomial(n+k-j,k-j))^2: j in [0..n]]) >; // Array
    A143007:= func< n,k | A(n-k,k) >; // Antidiagonal triangle
    [A143007(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 05 2023
    
  • Maple
    with(combinat): T:= (n,k) -> add(binomial(n+j,2*j)*binomial(2*j,j)^2*binomial(k+j,2*j), j = 0..n): for n from 0 to 9 do seq(T(n,k),k = 0..9) end do;
  • Mathematica
    T[n_, k_]:= HypergeometricPFQ[{-k, k+1, -n, n+1}, {1, 1, 1}, 1]; Table[T[n-k, k], {n,0,12}, {k,0,n}]//Flatten (* Jean-François Alcover, Mar 06 2013 *)
  • PARI
    /* Print as a square array: */
    {T(n, k)=sum(j=0, n, binomial(n+j, 2*j)*binomial(2*j, j)^2*binomial(k+j, 2*j))}
    for(n=0, 10, for(k=0,10, print1(T(n,k), ", "));print(""))
    
  • PARI
    /* (1) G.f. A(x,y) when read as a triangle: */
    {T(n,k)=local(A=1+x); A=sum(m=0, n, x^m * y^m / (1-x +x*O(x^n))^(2*m+1) * sum(k=0, m, binomial(m, k)^2*x^k)^2 ); polcoeff(polcoeff(A, n,x), k,y)}
    for(n=0, 10, for(k=0,n, print1(T(n,k), ", "));print(""))
    
  • PARI
    /* (2) G.f. A(x,y) when read as a triangle: */
    {T(n,k)=local(A=1+x); A=sum(m=0, n, x^m/(1-x*y +x*O(x^n))^(2*m+1) * sum(k=0, m, binomial(m, k)^2 * x^k * y^k)^2 ); polcoeff(polcoeff(A, n,x), k,y)}
    for(n=0, 10, for(k=0,n, print1(T(n,k), ", "));print(""))
    
  • PARI
    /* (3) G.f. A(x,y) when read as a triangle: */
    {T(n,k)=local(A=1+x); A=sum(m=0, n, x^m*sum(k=0, m, binomial(m , k)^2 * y^k * sum(j=0, k, binomial(k, j)^2 * x^j)+x*O(x^n))); polcoeff(polcoeff(A, n,x), k,y)}
    for(n=0, 10, for(k=0,n, print1(T(n,k), ", "));print(""))
    
  • PARI
    /* (4) G.f. A(x,y) when read as a triangle: */
    {T(n,k)=local(A=1+x); A=sum(m=0, n, x^m*sum(k=0, m, binomial(m, k)^2 * y^(m-k) * sum(j=0, k, binomial(k, j)^2 * x^j * y^j)+x*O(x^n))); polcoeff(polcoeff(A, n,x), k,y)}
    for(n=0, 10, for(k=0,n, print1(T(n,k), ", "));print(""))
    /* End */
    
  • SageMath
    def A(n,k): return sum((binomial(n,j)*binomial(n+k-j,k-j))^2 for j in range(n+1)) # array
    def A143007(n,k): return A(n-k,k) # antidiagonal triangle
    flatten([[A143007(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 05 2023

Formula

T(n,k) = Sum_{j = 0..n} C(n+j,2*j)*C(2*j,j)^2*C(k+j,2*j).
The array is symmetric T(n,k) = T(k,n).
The main diagonal [1,5,73,1445,...] is the sequence of Apery numbers A005259.
The entries in the k-th column satisfy the Apery-like recursion n^3*T(n,k) + (n-1)^3*T(n-2,k) = (2*n-1)*(n^2-n+1+2*k^2+2*k)*T(n-1,k).
The LDU factorization of the square array is L * D * transpose(L), where L is the lower triangular array A085478 and D is the diagonal matrix diag(C(2n,n)^2). O.g.f. for row n: The generating function for the coordination sequence of the lattice A_n is [Sum_{k = 0..n} C(n,k)^2*x^k ]/(1-x)^n. Thus the generating function for the coordination sequence of the product lattice A_n x A_n is {[Sum_{k = 0..n} C(n,k)^2*x^k]/(1-x)^n}^2 and hence the generating function for row n of this array, the crystal ball sequence of the lattice A_n x A_n, equals [Sum_{k = 0..n} C(n,k)^2*x^k]^2/(1-x)^(2n+1) = 1/(1-x)*[Legendre_P(n,(1+x)/(1-x))]^2. See [Conway & Sloane].
Series acceleration formulas for zeta(3): Row n: zeta(3) = (1 + 1/2^3 + ... + 1/n^3) + Sum_{k >= 1} 1/(k^3*T(n,k-1)*T(n,k)), n = 0,1,2,... . For example, the fourth row of the table (n = 3) gives zeta(3) = (1 + 1/2^3 + 1/3^3) + 1/(1^3*1*25) + 1/(2^3*25*253) + 1/(3^3*253*1445) + ... . See A143003 for further details.
Main diagonal: zeta(3) = 6 * Sum_{n >= 1} 1/(n^3*T(n-1,n-1)*T(n,n)). Conjectural result for other diagonals: zeta(3) = 1 + 1/2^3 + ... + 1/k^3 + Sum_{n >= 1} (2*n+k)*(3*n^2+3*n*k+k^2)/(n^3*(n+k)^3*T(n-1,n+k-1)*T(n,n+k)).
Sum_{k=0..n} T(n-k,k) = A227845(n) (antidiagonal sums). - Paul D. Hanna, Aug 27 2014
The main superdiagonal numbers S(n) := T(n,n+1) appear to satisfy the supercongruences S(m*p^r - 1) == S(m*p^(r-1) - 1) (mod p^(3*r)) for prime p >= 5 and m, r in N (this is true: see A352653. - Peter Bala, Apr 16 2022).
From Paul D. Hanna, Aug 27 2014: (Start)
G.f. A(x,y) = Sum_{n>=0, k=0..n} T(n,k)*x^n*y^k can be expressed by:
(1) Sum_{n>=0} x^n * y^n / (1-x)^(2*n+1) * [Sum_{k=0..n} C(n,k)^2 * x^k]^2,
(2) Sum_{n>=0} x^n / (1 - x*y)^(2*n+1) * [Sum_{k=0..n} C(n,k)^2 * x^k * y^k]^2,
(3) Sum_{n>=0} x^n * Sum_{k=0..n} C(n,k)^2 * y^k * Sum_{j=0..k} C(k,j)^2 * x^j,
(4) Sum_{n>=0} x^n * Sum_{k=0..n} C(n,k)^2 * y^(n-k) * Sum_{j=0..k} C(k,j)^2 * x^j * y^j. (End)
From Peter Bala, Jun 23 2023: (Start)
T(n,k) = Sum_{j = 0..n} C(n,j)^2 * C(n+k-j,k-j)^2.
T(n,k) = binomial(n+k,k)^2 * hypergeom([-n, -n, -k, -k],[-n - k, -n - k, 1], 1).
T(n,k) = hypergeom([n+1, -n, k+1, -k], [1, 1, 1], 1). (End)
From Peter Bala, Jun 28 2023: (Start)
T(n,k) = the coefficient of (x*z)^n*(y*t)^k in the expansion of 1/( (1 - x - y)*(1 - z - t) - x*y*z*t ).
T(n,k) = A(n, k, n, k) in the notation of Straub, equation 7.
The supercongruences T(n*p^r, k*p^r) == T(n*p^(r-1), k*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and positive integers n and k.
The formula T(n,k) = hypergeom([n+1, -n, k+1, -k], [1, 1, 1], 1) allows the table indexing to be extended to negative values of n and k; we have T(-n,k) = T(n-1,k) and T(n,-k) = T(n,k-1) leading to T(-n,-k) = T(n-1, k-1). (End)
From G. C. Greubel, Oct 05 2023: (Start)
Let t(n, k) = T(n-k, k) be the antidiagonal triangle, then:
t(n, k) = t(n, n-k).
Sum_{k=0..floor(n/2)} t(n-k,k) = A246563(n).
t(2*n+1, n+1) = A352653(n+1). (End)
From Peter Bala, Sep 27 2024: (Start)
The square array = A063007 * transpose(A063007) (LU factorization).
Let L denote the lower triangular array (l(n,k))n,k >= 0, where l(n, k) = (-1)^(n+k) * binomial(n, k)*binomial(n+k, k). (L is a signed version of A063007 and L = A063007 * A007318 ^(-1).)
Then the square array = L * transpose(A108625).
L^2 * transpose(A108625) = the Hadamard product of A108625 with itself (both identities can be verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package to find recurrences for the double sums involved). (End)

Extensions

Spelling/notation corrections by Charles R Greathouse IV, Mar 18 2010

A143003 a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+5)*a(n) - n^6*a(n-1).

Original entry on oeis.org

0, 1, 21, 1091, 114520, 21298264, 6410456640, 2923097201856, 1920450126458880, 1747596822651334656, 2133806329230225408000, 3405545462439659704320000, 6950705677729940374290432000, 17807686090745585163974737920000
Offset: 0

Views

Author

Peter Bala, Jul 19 2008

Keywords

Comments

This is the case m = 1 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+2*m^2+2*m+1)*a(n) - n^6*a(n-1) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series Sum_{k>=1} 1/k^3 for Apery's constant zeta(3). For other cases see A066989 (m=0), A143004 (m=2), A143005 (m=3) and A143006 (m=4).
The solution to the general recurrence may be expressed as a sum: a(n) = n!^3*p_m(n)*Sum_{k = 1..n} 1/(k^3*p_m(k-1)*p_m(k)), where p_m(x) = Sum_{k = 0..n} C(2*k,k)^2*C(n+k,2*k)*C(x+k,2*k) is a polynomial in x of degree 2*m.
The first few are p_0(x) = 1, p_1(x) = 2*x^2 + 2*x + 1, p_2(x) = (3*x^4 + 6*x^3 + 9*x^2 + 6*x + 2)/2 and p_3(x) = (10*x^6 + 30*x^5 + 85*x^4 + 120*x^3 + 121*x^2 + 66*x + 18)/18. For fixed n, the sequence [p_n(k)]k>=0 is the crystal ball sequence for the product lattice A_n x A_n. See A143007 for the table of values [p_n(k)] n,k >= 0. Observe that [p_n(n)] n >= 0 is the sequence of Apery numbers A005259.
The reciprocity law p_m(n) = p_n(m) holds for nonnegative integers m and n. In particular we have p_m(1) = 2*m^2 + 2*m + 1 and p_m(2) = (3*m^4 + 6*m^3 + 9*m^2 + 6*m + 2)/2.
The polynomial p_m(x) is the unique polynomial solution of the difference equation (x+1)^3*f(x+1) + x^3*f(x-1) = (2*x+1)*(x^2+x+2*m^2+2*m+1)*f(x), normalized so that f(0) = 1. The reciprocity law now yields the Apery-like recursion m^3*p_m(x) + (m-1)^3*p_(m-2)(x) = (2*m-1)*(m^2-m+1+2*x^2+2*x)*p_(m-1)(x).
The polynomial functions p_m(x) have their zeros on the vertical line Re x = -1/2 in the complex plane; that is, the polynomials p_m(x-1), m = 1,2,3,..., satisfy a Riemann hypothesis (adapt the proof of the lemma on p. 4 of [BUMP et al.]).
The general recurrence in the first paragraph above has a second solution b(n) = n!^3*p_m(n) with initial conditions b(0) = 1, b(1) = 2*m^2+2*m+1. Hence the behavior of a(n) for large n is given by lim_{n -> infinity} a(n)/b(n) = Sum_{k>=1} 1/(k^3*p_m(k-1)*p_m(k)) = 1/((2*m^2+2*m+1) - 1^6/(3*(2*m^2+2*m+3) - 2^6/(5*(2*m^2+2*m+7) - 3^6/(7*(2*m^2+2*m+13) - ...)))) = Sum_{k>=1} 1/(m+k)^3. The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Entry 32(iii)].
For the corresponding results for the constant zeta(2) see A142995. For corresponding results for the constant log(2) see A142979 and A142992.

References

  • Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

Crossrefs

The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)

Programs

  • Maple
    p := n -> 2*n^2+2*n+1: a := n -> n!^3*p(n)*sum (1/(k^3*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..14)
  • Mathematica
    RecurrenceTable[{a[0]==0,a[1]==1,a[n+1]==(2n+1)(n^2+n+5)a[n]- n^6 a[n-1]}, a[n],{n,15}] (* Harvey P. Dale, Jun 20 2011 *)

Formula

a(n) = n!^3*p(n)*Sum_{k = 1..n} 1/(k^3*p(k-1)*p(k)), where p(n) = 2*n^2 + 2*n + 1 = A001844(n).
Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+5)*a(n) - n^6*a(n-1).
The sequence b(n):= n!^3*p(n) satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 5. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(5 - 1^6/(21 - 2^6/(55 - 3^6/(119 - ... - (n-1)^6/((2*n-1)*(n^2-n+5)))))), for n >= 2. The behavior of a(n) for large n is given by lim_{n -> infinity} a(n)/b(n) = Sum_{k>=1} 1/(k^3*(4*k^4 + 1)) = 1/(5 - 1^6/(21 - 2^6/(55 - 3^6/(119 - ... - n^6/((2*n+1)*(n^2+n+5) - ...))))) = zeta(3) - 1, where the final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Entry 32(iii) at x = 1].

A066989 a(n) = (n!)^3 * Sum_{i=1..n} 1/i^3.

Original entry on oeis.org

1, 9, 251, 16280, 2048824, 444273984, 152759224512, 78340747014144, 57175952894078976, 57223737619918848000, 76212579497951858688000, 131758938842553681444864000, 289584291977410916858462208000, 794860754824699647616459210752000
Offset: 1

Views

Author

Benoit Cloitre, Jan 27 2002

Keywords

Comments

p^2 divides a(p-1) for prime p>5. - Alexander Adamchuk, Jul 11 2006

Crossrefs

Cf. A007408.
Column k=3 of A291556.

Programs

  • Mathematica
    f[k_] := k^3; t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[n - 1, t[n]]
    Table[a[n], {n, 1, 22}] (* A066989 *)
    (* Clark Kimberling, Dec 29 2011 *)
    Table[(n!)^3 * Sum[1/i^3, {i, 1, n}], {n, 1, 20}] (* Vaclav Kotesovec, Aug 27 2017 *)

Formula

Recurrence: a(1) = 1, a(2) = 9, a(n+2) = (2*n+3)*(n^2+3*n+3)*a(n+1) - (n+1)^6*a(n). b(n) = n!^3 satisfies the same recurrence with the initial conditions b(1) = 1, b(2) = 8. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(1-1^6/(9-2^6/(35-3^6/(91-...-(n-1)^6/((2n-1)*(n^2-n+1)))))) for n >= 2, leading to the infinite continued fraction expansion zeta(3) = 1/(1-1^6/(9-2^6/(35-3^6/(91-...-(n-1)^6/((2n-1)*(n^2-n+1)-...))))). Compare with A001819. - Peter Bala, Jul 19 2008
a(n) ~ Zeta(3) * (2*Pi)^(3/2) * n^(3*n+3/2) / exp(3*n). - Vaclav Kotesovec, Aug 27 2017
Sum_{n>=1} a(n) * x^n / (n!)^3 = polylog(3,x) / (1 - x). - Ilya Gutkovskiy, Jul 14 2020

A143004 a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+13)*a(n) - n^6*a(n-1).

Original entry on oeis.org

0, 1, 45, 4211, 704120, 191875384, 79755181632, 48072816950976, 40372248180436992, 45735898093934800896, 68049684624570789888000, 130036437291331549384704000, 313117351023401464093212672000
Offset: 0

Views

Author

Peter Bala, Jul 19 2008

Keywords

Comments

This is the case m = 2 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+2*m^2+2*m+1 )*a(n) - n^6*a(n-1) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series sum {k = 1..inf} 1/k^3 for Apery's constant zeta(3). For remarks on the general theory see A143003 (m=1). For other cases see A066989 (m=0), A143005 (m=3) and A143006 (m=4).

References

  • Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

Crossrefs

Programs

  • Maple
    p := n -> (3*n^4+6*n^3+9*n^2+6*n+2)/2: a := n -> n!^3*p(n)*sum (1/(k^3*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..14)
  • Mathematica
    RecurrenceTable[{a[0]==0,a[1]==1,a[n+1]==(2n+1)(n^2+n+13)a[n]-n^6 a[n-1]}, a,{n,20}] (* Harvey P. Dale, Jan 23 2012 *)

Formula

a(n) = n!^3*p(n)*sum {k = 1..n} 1/(k^3*p(k-1)*p(k)), where p(n) = (3*n^4+6*n^3+9*n^2+6*n+2)/2. Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+13)*a(n) - n^6*a(n-1). The sequence b(n):= n!^3*p(n) satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 13. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(13- 1^6/(45- 2^6/(95- 3^6/(175-...- (n-1)^6/((2*n-1)*(n^2-n+13)))))), for n >=2. The behavior of a(n) for large n is given by lim n -> infinity a(n)/b(n) = sum {k = 1..inf} 1/(k^3*p(k-1)*p(k)) = 1/(13- 1^6/(45- 2^6/(95- 3^6/(175-...- n^6/((2*n+1)*(n^2+n+13)-...))))) = zeta(3) - (1 + 1/2^3), where the final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Entry 32(iii) at x = 2].

A143006 a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+41)*a(n) - n^6*a(n-1).

Original entry on oeis.org

0, 1, 129, 30251, 11129080, 5985956824, 4501140404544, 4577449694922432, 6130634638657941504, 10576998168036159614976, 23068077837238986190848000, 62571876653848665596203008000, 208106990164246904792046895104000
Offset: 0

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Author

Peter Bala, Jul 19 2008

Keywords

Comments

This is the case m = 4 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+2*m^2+2*m+1 )*a(n) - n^6*a(n-1) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series sum {k = 1..inf} 1/k^3 for the constant zeta(3). For remarks on the general theory see A143003 (m=1). For other cases see A066989 (m=0), A143004 (m=2) and A143005 (m=3).

References

  • Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

Crossrefs

Programs

  • Maple
    p := n -> (35*n^8 +140*n^7 +630*n^6 +1400*n^5 +2595*n^4 +3020*n^3 +2500*n^2 +1200*n +288)/288: a := n -> n!^3*p(n)*sum (1/(k^3*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..13);

Formula

a(n) = n!^3*p(n)*sum {k = 1..n} 1/(k^3*p(k-1)*p(k)), where p(n) = (35*n^8 +140*n^7 +630*n^6 +1400*n^5 +2595*n^4 +3020*n^3 +2500*n^2 +1200*n +288)/288. Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+41)*a(n) - n^6*a(n-1). The sequence b(n):= n!^3*p(n) satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 41. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(41- 1^6/(129- 2^6/(235- 3^6/(371-...- (n-1)^6/((2*n-1)*(n^2-n+41)))))), for n >=2. The behavior of a(n) for large n is given by lim n -> infinity a(n)/b(n) = sum {k = 1..inf} 1/(k^3*p(k-1)*p(k)) = 1/(41- 1^6/(129- 2^6/(235- 3^6/(371-...- n^6/((2*n+1)*(n^2+n+41)-...))))) = zeta(3) - (1 + 1/2^3 + 1/3^3 + 1/4^3), where the final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Entry 32(iii) at x = 4].
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