A143003 -id:A143003 - cp's OEIS frontend

A143415 Another sequence of Apery-like numbers for the constant 1/e: a(n) = 1/(n+1)!Sum_{k = 0..n-1} C(n-1,k)(2*n-k)!.

0, 1, 5, 41, 481, 7421, 142601, 3288205, 88577021, 2731868921, 94969529101, 3675200329841, 156725471006105, 7302990263511541, 369216917569411601, 20130327811188977621, 1177435382675193700021, 73546210385434763486705

Offset: 0

Peter Bala, Aug 14 2008

This sequence is a modified version of A143414.

Seiichi Manyama, Table of n, a(n) for n = 0..366

Cf. A143413, A143414.

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.)

a := n -> 1/(n+1)!*add (binomial(n-1,k)*(2*n-k)!,k = 0..n-1): seq(a(n),n = 0..19);
# Alternative:
A143415 := n -> `if`(n=0, 0, ((2*n)!/(n+1)!)*hypergeom([1-n], [-2*n], 1)):
seq(simplify(A143415(n)), n = 0..17); # Peter Luschny, May 14 2020

Table[(1/(n+1)!)*Sum[Binomial[n-1,k]*(2*n-k)!, {k,0,n-1}], {n,0,50}] (* G. C. Greubel, Oct 24 2017 *)

for(n=0,25, print1((1/(n+1)!)*sum(k=0,n-1, binomial(n-1,k)*(2*n-k)!), ", ")) \\ G. C. Greubel, Oct 24 2017

a(n) = 1/(n+1)!*sum {k = 0..n-1} C(n-1,k)*(2*n-k)!.
a(n) = 1/(n*(n+1))*A143414(n) for n > 0.
Recurrence relation: a(0) = 0, a(1) = 1, (n-1)*(n+1)*a(n) - (n-2)*n*a(n-2) = (2*n-1)*(2*n^2-2*n+1)*a(n-1) for n >= 2. 1/e = 1/2 - 2 * Sum_{n = 1..inf} (-1)^(n+1)/(n*(n+2)*a(n)*a(n+1)) = 1/2 - 2*[1/(3*1*5) - 1/(8*5*41) + 1/(15*41*481) - 1/(24*481*7421) + ...] .
Conjectural congruences: for r >= 0 and prime p, calculation suggests the congruences a(p^r*(p+1)) == a(p^r) (mod p^(r+1)) may hold.
a(n) = ((2*n)!/(n+1)!)*hypergeom([1-n], [-2*n], 1) for n > 0. - Peter Luschny, May 14 2020

A219692 a(n) = Sum_{j=0..floor(n/3)} (-1)^j C(n,j) * C(2j,j) * C(2n-2j,n-j) * (C(2n-3j-1,n) + C(2n-3j,n)).

Original entry on oeis.org

2, 6, 54, 564, 6390, 76356, 948276, 12132504, 158984694, 2124923460, 28877309604, 398046897144, 5554209125556, 78328566695736, 1114923122685720, 15999482238880464, 231253045986317814, 3363838379489630916

Offset: 0

Jason Kimberley, Nov 25 2012

This sequence is s_18 in Cooper's paper.
This is one of the Apery-like sequences - see Cross-references. - Hugo Pfoertner, Aug 06 2017
Every prime eventually divides some term of this sequence. - Amita Malik, Aug 20 2017

G. C. Greubel, Table of n, a(n) for n = 0..830 (terms 0..254 from Jason Kimberley)
S. Cooper, Sporadic sequences, modular forms and new series for 1/pi, Ramanujan J. (2012).
Ofir Gorodetsky, New representations for all sporadic Apéry-like sequences, with applications to congruences, arXiv:2102.11839 [math.NT], 2021. See s18 p. 3.
Amita Malik and Armin Straub, Divisibility properties of sporadic Apéry-like numbers, Research in Number Theory, 2016, 2:5.

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.)

Magma
```
s_18 := func where C is Binomial;
```

Table[Sum[(-1)^j*Binomial[n,j]*Binomial[2j,j]*Binomial[2n-2j, n-j]* (Binomial[2n-3j-1,n] +Binomial[2n-3j,n]), {j,0,Floor[n/3]}], {n,0,20}] (* G. C. Greubel, Oct 24 2017 *)

{a(n) = sum(j=0,floor(n/3), (-1)^j*binomial(n,j)*binomial(2*j,j)* binomial(2*n-2*j,n-j)*(binomial(2*n-3*j-1,n) +binomial(2*n-3*j,n)))}; \\ G. C. Greubel, Apr 02 2019

[sum((-1)^j*binomial(n,j)*binomial(2*j,j)*binomial(2*n-2*j,n-j)* (binomial(2*n-3*j-1,n)+binomial(2*n-3*j,n)) for j in (0..floor(n/3))) for n in (0..20)] # G. C. Greubel, Apr 02 2019

1/Pi
= 2*3^(-5/2) Sum {k>=0} (n a(n)/18^n) [Cooper, equation (42)]
= 2*3^(-5/2) Sum {k>=0} (n a(n)/A001027(n)).
G.f.: 1+hypergeom([1/8, 3/8],[1],256*x^3/(1-12*x)^2)^2/sqrt(1-12*x). - Mark van Hoeij, May 07 2013
Conjecture D-finite with recurrence: n^3*a(n) -2*(2*n-1)*(7*n^2-7*n+3)*a(n-1) +12*(4*n-5)*(n-1)* (4*n-3)*a(n-2)=0. - R. J. Mathar, Jun 14 2016
a(n) ~ 3 * 2^(4*n + 1/2) / (Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Mar 08 2023

A260667 a(n) = (1/n^2) * Sum_{k=0..n-1} (2k+1)S(k,n)^2, where S(k,x) denotes the polynomial Sum_{j=0..k} binomial(k,j)binomial(x,j)*binomial(x+j,j).

Original entry on oeis.org

1, 37, 1737, 102501, 6979833, 523680739, 42129659113, 3572184623653, 315561396741609, 28807571694394593, 2701627814373536601, 259121323945378645947, 25330657454041707496017, 2516984276442279642274311, 253667099464270541534450025, 25884030861250181046253181349, 2670255662315910532447096232073

Offset: 1

Zhi-Wei Sun, Nov 14 2015

nonn

Conjecture: For k = 0,1,2,... define S(k,x):= Sum_{j=0..k} binomial(k,j)*binomial(x,j)*binomial(x+j,j).
(i) For any integer n > 0, the polynomial (1/n^2) * Sum_{k=0..n-1}(2k+1)*S(k,x)^2 is integer-valued (and hence a(n) is always integral).
(ii) Let r be 0 or 1, and let x be any integer. Then, for any positive integers m and n, we have the congruence
   Sum_{k=0..n-1} (-1)^(k*r)*(2k+1)*S(k,x)^(2m) == 0 (mod n).
(iii) For any odd prime p, we have Sum_{k=0..p-1} S(k,-1/2)^2 == (-1/p)(1-7*p^3*B_{p-3}) (mod p^4), where (a/p) is the Legendre symbol, and B_0,B_1,B_2,... are Bernoulli numbers. Also, for any prime p > 3 we have Sum_{k=0..p-1} S(k,-1/3)^2 == p - (14/3)*(p/3)*p^3*B_{p-2}(1/3) (mod p^4), where B_n(x) denotes the Bernoulli polynomial of degree n; Sum_{k=0..p-1} S(k,-1/4)^2 == (2/p)*p - 26*(-2/p)*p^3*E_{p-3} (mod p^4), where E_0,E_1,E_2,... are Euler numbers; Sum_{k=0..p-1} S(k,-1/6)^2 == (3/p)*p - (155/12)*(-1/p)*p^3*B_{p-2}(1/3) (mod p^4).
Our conjecture is motivated by a conjecture of Kimoto and Wakayama which states that Sum_{k=0..p-1} S(k,-1/2)^2 == (-1/p) (mod p^3) for any odd prime p. The Kimoto-Wakayama conjecture was confirmed by Long, Osburn and Swisher in 2014.
For more related conjectures, see Sun's paper arXiv.1512.00712. - Zhi-Wei Sun, Dec 03 2015

			a(2) = 37 since (1/2^2) * Sum_{k=0..1} (2k+1)*S(k,2)^2 = (S(0,2)^2 + 3*S(1,2)^2)/4 = (1^2 + 3*7^2)/4 = 148/4 = 37.
G.f. = x + 37*x^2 + 1737*x^3 + 102501*x^4 + 6979833*x^5 + 523680739*x^6 + ...

Zhi-Wei Sun, Table of n, a(n) for n = 1..100
K. Kimoto and M. Wakayama, Apéry-like numbers arising from special values of spectral zeta function for non-commutative harmonic oscillators, Kyushu J. Math. 60(2006), no.2, 383-404. (Cf. the formula (6.19).)
L. Long, R. Osburn and H. Swisher, On a conjecture of Kimoto and Wakayama, arXiv:1404.4723 [math.NT], 2014.
Z.-W. Sun, Supercongruences involving products of two binomial coefficients, arXiv:1011.6676 [math.NT], 2010-2013; Finite Fields Appl. 22(2013), 24-44.
Z.-W. Sun, Congruences involving g_n(x)=sum_{k=0..n} binomial(n,k)^2*binomial(2k,k)*x^k, Ramanujan J., in press. Doi: 10.1007/s11139-015-9727-3.
Z.-W. Sun, Supercongruences involving dual sequences, arXiv:1502.00712 [math.NT], 2015.

Cf. A000290, A027641, A027642, A122045.

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.)

# Implementing Mark van Hoeij's formula.
c := n -> binomial(2*n, n)/(n + 1):
h := n -> simplify(hypergeom([-n,-n,-n], [1,-2*n], 1)):
b := n -> c(n)^2*((n+11)*(2+4*n)^2*h(n+1)^2-2*(n+1)*(11*n+16)*(1+2*n)*h(n)*h(n+1)-h(n)^2*(n+1)^3)/(25*(n+2)):
a := n -> b(n-1): seq(a(n), n = 1..17);  # Peter Luschny, Nov 11 2022

S[k_,x_]:=S[k,x]=Sum[Binomial[k,j]Binomial[x,j]Binomial[x+j,j],{j,0,k}]
a[n_]:=a[n]=Sum[(2k+1)*S[k,n]^2,{k,0,n-1}]/n^2
Do[Print[n," ",a[n]],{n,1,17}]

a(n) ~ phi^(10*n + 3) / (10 * Pi^2 * n^3), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Nov 06 2021
Conjecture: a(p-1) == 1 (mod p^3) for all primes p >= 5. - Peter Bala, Aug 15 2022
a(n) = ((n+10)*A005258(n)^2 - (11*n+5)*A005258(n)*A005258(n-1) - n*A005258(n-1)^2)/(25*(n+1)). - Mark van Hoeij, Nov 11 2022

A262177 Decimal expansion of Q_5 = zeta(5) / (Sum_{k>=1} (-1)^(k+1) / (k^5 * binomial(2k, k))), a conjecturally irrational constant defined by an Apéry-like formula.

Original entry on oeis.org

2, 0, 9, 4, 8, 6, 8, 6, 2, 2, 0, 1, 0, 0, 3, 6, 9, 9, 3, 8, 5, 0, 2, 4, 9, 2, 9, 3, 7, 3, 2, 9, 4, 1, 6, 3, 0, 2, 9, 6, 7, 5, 8, 7, 4, 8, 5, 6, 7, 7, 8, 1, 8, 2, 7, 4, 0, 1, 2, 7, 5, 8, 7, 8, 3, 7, 4, 3, 8, 0, 0, 7, 8, 7, 6, 8, 4, 6, 8, 1, 5, 6, 3, 2, 0, 6, 0, 4, 4, 2, 3, 2, 0, 9, 0, 4, 3, 1, 3, 6, 9, 3, 1

Offset: 1

Jean-François Alcover, Sep 14 2015

The similar constant Q_3 = zeta(3) / (Sum_{k>=1} (-1)^(k+1) / (k^3 * binomial(2k, k))) evaluates to 5/2.

			2.09486862201003699385024929373294163029675874856778182740127587837438...

G. C. Greubel, Table of n, a(n) for n = 1..5000
David Bailey, Jonathan Borwein, David Bradley, Experimental determination of Apéry-like identities for zeta(2n+2), arXiv:math/0505270 [math.NT], 2005.

Cf. A013663.

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.)

Q5 = Zeta[5]/Sum[(-1)^(k+1)/(k^5*Binomial[2k, k]), {k, 1, Infinity}]; RealDigits[Q5, 10, 103] // First

zeta(5)/suminf(k=1, (-1)^(k+1)/(k^5*binomial(2*k,k))) \\ Michel Marcus, Sep 14 2015

Equals 2*zeta(5)/6F5(1,1,1,1,1,1; 3/2,2,2,2,2; -1/4).

A142979 a(1) = 1, a(2) = 3, a(n+2) = 3a(n+1) + (n+1)^2a(n).

Original entry on oeis.org

1, 3, 13, 66, 406, 2868, 23220, 210192, 2116656, 23375520, 281792160, 3673814400, 51599514240, 775673176320, 12440524320000, 211848037632000, 3820318338816000, 72685037892096000, 1455838255452672000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 1 of the more general recurrence a(1) = 1, a(2) = 2*m + 1, a(n+2) = (2*m + 1)*a(n+1) + (n + 1)^2*a(n) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of Mercator's series for the constant log(2). For other cases see A024167 (m = 0), A142980 (m = 2), A142981 (m = 3) and A142982 (m = 4).
The solution to the general recurrence may be expressed as a sum: a(n) = n!*p_m(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p_m(k-1)*p_m(k)), where p_m(x) = Sum_{k = 0..m} 2^k*C(m,k)*C(x,k) = Sum_{k = 0..m} C(m,k)*C(x+k,m), is the Ehrhart polynomial of the m-dimensional cross polytope (the hyperoctahedron).
The first few values are p_0(x) = 1, p_1(x) = 2*x + 1, p_2(x) = 2*x^2 + 2*x + 1 and p_3(x) = (4*x^3 + 6*x^2 + 8*x + 3)/3.
The sequence {p_m(k)},k>=0 is the crystal ball sequence for the product lattice A_1 x... x A_1 (m copies). The table of values [p_m(k)]m,k>=0 is the array of Delannoy numbers A008288.
The polynomial p_m(x) is the unique polynomial solution of the difference equation (x + 1)*f(x+1) - x*f(x-1) = (2*m + 1)*f(x), normalized so that f(0) = 1. These polynomials 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 [BUMP et al., Theorems 4 and 6]. The o.g.f. for the p_m(x) is (1 + t)^x/(1 - t)^(x+1) = 1 + (2*x + 1)*t + (2*x^2 + 2*x + 1)*t^2 + ....
The general recurrence in the first paragraph above also has a second solution b(n) = n!*p_m(n) with initial conditions b(1) = 2*m + 1, b(2) = (2*m + 1)^2 + 1.
Hence the behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(k*p_m(k-1)*p_m(k)) = 1/((2*m + 1) + 1^2/((2*m + 1) + 2^2/((2*m + 1) + 3^2/((2*m + 1) + ... + n^2/((2*m + 1) + ...))))) = (-1)^m * (log(2) - (1 - 1/2 + 1/3 - ... + (-1)^(m+1)/m)), where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 29]).
For other sequences defined by similar recurrences and related to log(2) see A142983 and A142988. See also A142992 for the connection between log(2) and the C_n lattices. For corresponding results for the constants e, zeta(2) and zeta(3) see A000522, A142995 and A143003 respectively.

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

Seiichi Manyama, Table of n, a(n) for n = 1..448
D. Bump, K. Choi, P. Kurlberg and J. Vaaler, A local Riemann hypothesis, I, Math. Zeit. 233, (2000), 1-19.

Cf. A008288, A024167, A142980, A142981, A142982, A142983, A142988, A142992, A317057.

p := n -> 2*n+1: a := n -> n!*p(n)*sum ((-1)^(k+1)/(k*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 1..20)

RecurrenceTable[{a[1]==1,a[2]==3,a[n+2]==3a[n+1]+(n+1)^2 a[n]},a,{n,20}] (* Harvey P. Dale, May 20 2012 *)

a(n) = n!*p(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p(k-1)*p(k)), where p(n) = 2*n + 1.
Recurrence: a(1) = 1, a(2) = 3, a(n+2) = 3*a(n+1) + (n + 1)^2*a(n).
The sequence b(n):= n!*p(n) satisfies the same recurrence with b(1) = 3, b(2) = 10.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(3 + 1^2/(3 + 2^2/(3 + 3^2/(3 + ... + (n-1)^2/3)))), for n >= 2.
Limit_{n -> oo} a(n)/b(n) = 1/(3 + 1^2/(3 + 2^2/(3 + 3^2/(3 + ... + (n-1)^2/(3 + ...))))) = Sum_{k >= 1} (-1)^(k+1)/(k*(4k^2 - 1)) = 1 - log(2).
Thus a(n) ~ c*n*n! as n -> oo, where c = 2*(1 - log(2)).
From Peter Bala, Dec 09 2024: (Start)
E.g.f.: A(x) = (2*x - (1 + x)*log(1 + x))/(1 - x)^2 satisfies the differential equation 1 + (x + 3)*A(x) + (x^2 - 1)*A'(x) = 0 with A(0) = 0.
Sum_{k = 1..n} Stirling_2(n, k) * a(k) = A317057(n+1). (End)

A142983 a(1) = 1, a(2) = 2, a(n+2) = 2a(n+1) + (n + 1)(n + 2)*a(n).

Original entry on oeis.org

1, 2, 10, 44, 288, 1896, 15888, 137952, 1419840, 15255360, 186693120, 2387093760, 33898314240, 502247692800, 8123141376000, 136785729024000, 2483065912320000, 46822564905984000, 942853671825408000, 19678282007924736000, 435355106182520832000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 1 of the general recurrence a(1) = 1, a(2) = 2*m, a(n+2) = 2*m*a(n+1) + (n + 1)*(n + 2)*a(n) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series 1/2 + 1/2*Sum_{k > 1} (-1)^(k+1)/(k*(k + 1)) = log(2). For other cases see A142984 (m = 2), A142985 (m = 3), A142986 (m = 4) and A142987 (m = 5).
The solution to the general recurrence may be expressed as a sum: a(n) = n!*p_m(n+1)*Sum_{k = 1..n} (-1)^(k+1)/(p_m(k)*p_m(k+1)), where p_m(x) = Sum_{k = 1..m} 2^(k-1)*C(m-1,k-1)*C(x,k) is the polynomial that gives the regular polytope numbers for the m-dimensional cross polytope as defined by [Kim] (see A142978). The first few values are p_1(x) = x, p_2(x) = x^2, p_3(x) = (2*x^3 + x)/3 and p_4(x) = (x^4 + 2*x^2)/3.
The polynomial p_m(x) is the unique polynomial solution of the difference equation x*(f(x+1) - f(x-1)) = 2*m*f(x), normalized so that f(1) = 1.
The o.g.f. for the p_m(x) is 1/2*((1 + t)/(1 - t))^x = 1/2 + x*t + x^2*t^2 + (2*x^3 + x)/3*t^3 + .... Thus p_m(x) is, apart from a constant factor, the Meixner polynomial of the first kind M_m(x;b,c) at b = 0, c = -1, also known as a Mittag-Leffler polynomial.
The general recurrence in the first paragraph above has a second solution b(n) = n!*p_m(n+1) with b(1) = 2*m, b(2) = m^2 + 2. Hence the behavior of a(n) for large n is given by Limit_{n-> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(p_m(k)*p_m(k+1)) = 1/((2*m) + 1*2/((2*m) + 2*3/((2*m) + 3*4/((2*m) + ... + n*(n + 1)/((2*m) + ...))))) = 1 + (-1)^(m+1) * (2*m)*(log(2) - (1 - 1/2 + 1/3 - ... + (-1)^(m+1)/m)), where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 32(i)]).
See A142979, A142988 and A142992 for similar results. For corresponding results for Napier's constant e, the constant zeta(2) and Apery's constant zeta(3) refer to A000522, A142995 and A143003, respectively.

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

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2002), 65-75.
Eric Weisstein's World of Mathematics, Meixner polynomial of the first kind.
Eric Weisstein's World of Mathematics, Mittag-Leffler polynomial.

Cf. A000522, A142979, A142984, A142985, A142986, A142987, A142988, A142992.

Cf. A002378.

a142983 n = a142983_list !! (n-1)
a142983_list = 1 : 2 : zipWith (+)
                       (map (* 2) $ tail a142983_list)
                       (zipWith (*) (drop 2 a002378_list) a142983_list)
-- Reinhard Zumkeller, Jul 17 2015

a := n -> (n+1)!*sum ((-1)^(k+1)/(k*(k+1)), k = 1..n): seq(a(n), n = 1..20);

Rest[CoefficientList[Series[(-x+2*Log[x+1])/(x-1)^2,{x,0,20}],x]*Range[0,20]!] (* Vaclav Kotesovec, Oct 21 2012 *)

a(n) = n!*p(n+1)*Sum_{k = 1..n} (-1)^(k+1)/(p(k)*p(k+1)), where p(n) = n.
Recurrence: a(1) = 1, a(2) = 2, a(n+2) = 2*a(n+1) + (n + 1)*(n + 2)*a(n).
The sequence b(n) := n!*p(n+1) satisfies the same recurrence with b(1) = 2 and b(2) = 6.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(2 + 1*2/(2 + 2*3/(2 + 3*4/(2 + ... + (n - 1)*n/2)))), for n >= 2.
The behavior of a(n) for large n is given by Limit_{n -> oo} a(n)/b(n) = 1/(2 + 1*2/(2 + 2*3/(2 + 3*4/(2 + ... + n*(n+1)/(2 + ...))))) = Sum_{k >= 1} (-1)^(k+1)/(k*(k + 1)) = 2*log(2) - 1.
E.g.f.: (2*log(x+1)-x)/(x-1)^2. - Vaclav Kotesovec, Oct 21 2012

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

Benoit Cloitre, Jan 27 2002

nonn

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

Seiichi Manyama, Table of n, a(n) for n = 1..181 (terms 1..50 from T. D. Noe)

Cf. A007408.
Cf. A001819, A143003, A143004, A143005, A143006.
Column k=3 of A291556.

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 *)

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

A142995 a(0) = 0, a(1) = 1, a(n+1) = (2n^2 + 2n + 3)a(n) - n^4a(n-1), n >= 1.

Original entry on oeis.org

0, 1, 7, 89, 1836, 56164, 2390832, 135213840, 9809203968, 888117094656, 98167241088000, 13010123816064000, 2036436482119680000, 371699564417796096000, 78251077775510986752000

Offset: 0

Peter Bala, Jul 18 2008

This is the case m = 1 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n^2 + 2*n + m^2 + m + 1)*a(n) - n^4*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^2 for the constant zeta(2). For other cases see A001819 (m=0), A142996 (m=2), A142997 (m=3) and A142998 (m=4).
The solution to the general recurrence may be expressed as a sum: a(n) = n!^2*p_m(n)*Sum_{k = 1..n} 1/(k^2*p_m(k-1)*p_m(k)), where p_m(x) := Sum_{k = 0..m} C(m,k)^2*C(x+k,m) = Sum_{k = 0..m} C(m,k)*C(m+k,k)*C(x,k) is the Ehrhart polynomial of the polytope formed from the convex hull of a root system of type A_m (equivalently, the polynomial that generates the crystal ball sequence for the A_m lattice [Bacher et al.]).
The first few are p_0(x) = 1, p_1(x) = 2*x + 1, p_2(x) = 3*x^2 + 3*x + 1 and p_3(x) = (10*x^3 + 15*x^2 + 11*x + 3)/3. The o.g.f. for the p_m(x) is ((1-t^2)^x/(1-t)^(2x+1))*Legendre_P(x,(1+t^2)/(1-t^2)) = 1 + (2*x+1)*t + (3*x^2+3*x+1)*t^2 + ... [Gogin & Hirvensalo, Theorem 1 with N = -1].
The polynomial p_m(x) is the unique polynomial solution of the difference equation (x+1)^2*f(x+1) + x^2*f(x-1) = (2*x^2 + 2*x + m^2 + m + 1)*f(x), normalized so that f(0) = 1. These polynomials 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!^2*p_m(n) with initial conditions b(0) = 1, b(1) = m^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^2*p_m(k-1)*p_m(k)) = 1/((m^2 + m + 1) - 1^4/((m^2 + m + 5) - 2^4/((m^2 + m + 13) - ... - n^4/((2*n^2 + 2*n + m^2 + m + 1) - ...)))) = 2*Sum_{k>=1} (-1)^(k+1)/(m+k)^2. The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Corollary to Entry 31] (replace x by 2x+1 in the corollary and apply Entry 14).
For related results see A142999. For corresponding results for the constants e, log(2) and zeta(3) see A000522, A142979 and A143003 respectively.

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

Seiichi Manyama, Table of n, a(n) for n = 0..252
R. Bacher, P. de la Harpe and B. Venkov, Séries de croissance et séries d'Ehrhart associées aux réseaux de racines, C. R. Acad. Sci. Paris, 325 (Series 1) (1997), 1137-1142.
D. Bump, K. Choi, P. Kurlberg and J. Vaaler, A local Riemann hypothesis, I, Math. Zeit. 233, (2000), 1-19.
N. Gogin and M. Hirvensalo, On the generating function of discrete Chebyshev polynomials, Turku Centre for Computer Science Technical Report No. 819, (2007), 1-8.

Cf. A000522, A001819, A003215 (A_2 lattice), A005902 (A_3 lattice), A008384 (A_4 lattice), A008386 (A_5 lattice), A108625, A142979, A142996, A142997, A142998, A143003.

p := n -> 2*n+1: a := n -> n!^2*p(n)*sum (1/(k^2*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..20);

a[n_] := -1/6*n!^2*(2*n*(Pi^2-12) + Pi^2 - 6*(2*n+1)*PolyGamma[1, n+1]) // Simplify; Table[a[n], {n, 0, 14}] (* Jean-François Alcover, Mar 06 2013 *)

a(n) = n!^2*p(n)*Sum_{k = 1..n} 1/(k^2*p(k-1)*p(k)), where p(n) = 2*n+1. Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n^2 + 2*n + 3)*a(n) - n^4*a(n-1). The sequence b(n):= n!^2*p(n) satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 3. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(3 - 1^4/(7 - 2^4/(15 - 3^4/(27 - ... - (n-1)^4/(2*n^2 - 2*n + 3))))), for n >= 2. Lim_{n -> infinity} a(n)/b(n) = 1/(3 - 1^4/(7 - 2^4/(15 - 3^4/(27 - ... - n^4/((2*n^2 + 2*n + 3) - ...))))) = Sum_{k>=1} 1/(k^2*(4*k^2 - 1)) = 2 - zeta(2).

A142999 a(0) = 0, a(1) = 1; for n > 1, a(n+1) = (2n + 1)a(n) + n^4*a(n-1).

Original entry on oeis.org

0, 1, 3, 31, 460, 12076, 420336, 21114864, 1325949696, 109027627776, 10771080883200, 1316468976307200, 187978181665996800, 31997755234356019200, 6232784237890147123200, 1409976507981835100160000, 359243973790625586216960000, 104259271562188189469245440000

Offset: 0

Peter Bala, Jul 18 2008

This is the case m = 0 of the general recurrence a(0) = 1, a(1) = 1, a(n+1) = (2*m + 1)*(2*n + 1)*a(n) + n^4*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+1)/k^2 for the constant 1/2*zeta(2). For other cases see A143000 (m = 1), A143001 (m = 2) and A143002 (m = 3).
The solution to the general recurrence may be expressed as a sum: a(n) = n!^2*p_m(n)*Sum_{k = 1..n} (-1)^(k+1)/(k^2*p_m(k-1)*p_m(k)), where p_m(x) := Sum_ {k = 0..m} C(m,k)*C(x,k)*C(x+k,k). Note that the polynomial q_m(x) := Sum_{k = 0..m} C(m,k)*C(m+k,k)*C(x,k), obtained by interchanging the roles of m and x, may be variously described as the Ehrhart polynomial of the polytope formed from the convex hull of a root system of type A_m, the polynomial that generates the crystal ball sequence for the A_m lattice [Bacher et al.], or the discrete Chebyshev polynomial D_m(N;x) at N = -1 [Gogin & Hirvensalo]. Compare with the comments in A142995.
The first few values are p_0(x) = 1, p_1(x) = x^2 + x + 1, p_2(x) = (x^4 + 2*x^3 + 7*x^2 + 6*x + 4)/4 and p_3(x) = (x^6 + 3*x^5 + 22*x^4 + 39*x^3 + 85*x^2 + 66*x + 36)/36.
The polynomial p_m(x) is the unique polynomial solution of the difference equation (x + 1)^2*f(x+1) - x^2*f(x-1) = (2*m + 1)*(2*x + 1)*f(x), normalized so that f(0) = 1. These polynomials 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!^2)*p_m(n) with initial conditions b(0) = 1, b(1) = 2*m + 1. Hence the behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(k^2*p_m(k-1)*p_m(k)) = 1/((2*m + 1) + 1^4/(3*(2*m + 1) + 2^4/(5*(2*m + 1) + ... + n^4/(((2*n + 1)*(2*m + 1) + ...)))) = (1/2)*Sum_{k >= 1} 1/(m + k)^2. The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Corollary to Entry 30].
For results of a similar nature for the constants e, log(2), zeta(2) and zeta(3) see A000522, A142979, A142995 and A143003 respectively.

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

Vincenzo Librandi, Table of n, a(n) for n = 0..200
R. Bacher, P. de la Harpe and B. Venkov, Séries de croissance et polynômes d'Ehrhart associées aux réseaux de racines, Annales de l'Institut Fourier, Tome 49 (1999) no. 3, pp. 727-762.
R. Bacher, P. de la Harpe and B. Venkov, Séries de croissance et séries d'Ehrhart associées aux réseaux de racines, C. R. Acad. Sci. Paris, 325 (Series 1) (1997), 1137-1142.
D. Bump, K. Choi, P. Kurlberg and J. Vaaler, A local Riemann hypothesis, I, Math. Zeit. 233, (2000), 1-19.
N. Gogin and M. Hirvensalo, On the generating function of discrete Chebyshev polynomials, Turku Centre for Computer Science Technical Report No. 819, (2007), 1-8.

Cf. A000522, A142979, A142995, A143000, A143001, A143002, A143003.

a := n -> n!^2*add ((-1)^(k+1)/k^2, k = 1..n): seq(a(n), n = 0..20);

f[k_] := (k^2) (-1)^(k + 1)
t[n_] := Table[f[k], {k, 1, n}]
a[n_] := SymmetricPolynomial[n - 1, t[n]]
Table[a[n], {n, 1, 18}]    (* A142999 signed *)
(* Clark Kimberling, Dec 30 2011 *)
RecurrenceTable[{a[0]==0,a[1]==1,a[n]==(2(n-1)+1)a[n-1]+(n-1)^4 a[n-2]},a,{n,20}] (* Harvey P. Dale, Apr 26 2014 *)

a(n) = (n!^2) * Sum_{k = 1..n} (-1)^(k+1)/k^2.
Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n + 1)*a(n) + (n^4)*a(n-1).
The sequence b(n) := n!^2 satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 1. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(1 + 1^4/(3 + 2^4/(5 + 3^4/(7 + ... + (n - 1)^4/(2*n - 1))))), for n >= 2.
Limit_{n -> oo} a(n)/b(n) = 1/(1 + 1^4/(3 + 2^4/(5 + 3^4/(7 + ... + n^4/((2*n + 1) + ...))))) = Sum_{k >= 1} (-1)^(k+1)/k^2 = 1/2*zeta(2).
Sum_{n>=0} a(n) * x^n / (n!)^2 = -polylog(2,-x) / (1 - x). - Ilya Gutkovskiy, Jul 15 2020

a(0)=0 added by Vincenzo Librandi, Apr 27 2014

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

Original entry on oeis.org

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

Offset: 0

Peter Bala, Jul 19 2008

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).

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

Seiichi Manyama, Table of n, a(n) for n = 0..180

Cf. A066989, A143003, A143005, A143006, A143007.

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)

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 *)

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].

cp's OEIS Frontend

A143415 Another sequence of Apery-like numbers for the constant 1/e: a(n) = 1/(n+1)!*Sum_{k = 0..n-1} C(n-1,k)*(2*n-k)!.

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A219692 a(n) = Sum_{j=0..floor(n/3)} (-1)^j C(n,j) * C(2j,j) * C(2n-2j,n-j) * (C(2n-3j-1,n) + C(2n-3j,n)).

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A260667 a(n) = (1/n^2) * Sum_{k=0..n-1} (2k+1)*S(k,n)^2, where S(k,x) denotes the polynomial Sum_{j=0..k} binomial(k,j)*binomial(x,j)*binomial(x+j,j).

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A262177 Decimal expansion of Q_5 = zeta(5) / (Sum_{k>=1} (-1)^(k+1) / (k^5 * binomial(2k, k))), a conjecturally irrational constant defined by an Apéry-like formula.

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A142979 a(1) = 1, a(2) = 3, a(n+2) = 3*a(n+1) + (n+1)^2*a(n).

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A142983 a(1) = 1, a(2) = 2, a(n+2) = 2*a(n+1) + (n + 1)*(n + 2)*a(n).

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

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A143415 Another sequence of Apery-like numbers for the constant 1/e: a(n) = 1/(n+1)!Sum_{k = 0..n-1} C(n-1,k)(2*n-k)!.

A260667 a(n) = (1/n^2) * Sum_{k=0..n-1} (2k+1)S(k,n)^2, where S(k,x) denotes the polynomial Sum_{j=0..k} binomial(k,j)binomial(x,j)*binomial(x+j,j).

A142979 a(1) = 1, a(2) = 3, a(n+2) = 3a(n+1) + (n+1)^2a(n).

A142983 a(1) = 1, a(2) = 2, a(n+2) = 2a(n+1) + (n + 1)(n + 2)*a(n).

A142995 a(0) = 0, a(1) = 1, a(n+1) = (2n^2 + 2n + 3)a(n) - n^4a(n-1), n >= 1.

A142999 a(0) = 0, a(1) = 1; for n > 1, a(n+1) = (2n + 1)a(n) + n^4*a(n-1).

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