A142988 -id:A142988 - cp's OEIS frontend

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

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

A142989 a(1) = 1, a(2) = 5, a(n+2) = 5a(n+1)+(n+1)(n+3)*a(n).

Original entry on oeis.org

1, 5, 33, 240, 1992, 18360, 187416, 2093760, 25462080, 334592640, 4728412800, 71488811520, 1151817408000, 19699405286400, 356504125824000, 6805868977152000, 136702533123072000, 2881808345235456000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 1 of the general recurrence a(1) = 1, a(2) = 2*m+3, a(n+2) = (2*m+3)*a(n+1)+(n+1)*(n+3)*a(n), which arises when accelerating the convergence of a certain series for the constant log(2). For remarks on the general case see A142988 (m=0). For other cases see A142990 (m=2) and A142991 (m=3).

Harvey P. Dale, Table of n, a(n) for n = 1..447

Cf. A142979, A142983, A142988, A142990, A142991.

p := n -> (2*n-1)/3: a := n -> (n+2)!*p(n+2)*sum ((-1)^(k+1)/(k*(k+1)*(k+2)*p(k+1)*p(k+2)), k = 1..n): seq(a(n), n = 1..20);

RecurrenceTable[{a[1]==1,a[2]==5,a[n]==5a[n-1]+(n-1)(n+1)a[n-2]},a,{n,20}] (* Harvey P. Dale, Jun 17 2013 *)

a(n) = (n+2)!*p(n+2)*sum {k = 1..n} (-1)^(k+1)/(k*(k+1)*(k+2)*p(k+1)*p(k+2)), where p(n) = (2*n-1)/3. Recurrence: a(1) = 1, a(2) = 5, a(n+2) = 5*a(n+1)+(n+1)*(n+3)*a(n). The sequence b(n) := 1/2*(n+2)!*p(n+2) satisfies the same recurrence with b(1) = 5, b(2) = 28. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(5+1*3/(5+2*4/(5+3*5/(5+...+(n-1)*(n+1)/5)))), for n >=2. Lim n -> infinity a(n)/b(n) = 1/(5+1*3/(5+2*4/(5+3*5/(5+...+(n-1)*(n+1)/(5+...))))) = 2*sum {k = 1..inf} (-1)^(k+1)/ (k*(k+1)*(k+2)*p(k+1)*p(k+2)) = 17/2-12*log(2).

A142990 a(1) = 1, a(2) = 7, a(n+2) = 7a(n+1)+(n+1)(n+3)*a(n).

Original entry on oeis.org

1, 7, 57, 504, 4896, 51912, 598392, 7459200, 100085760, 1439061120, 22083719040, 360371773440, 6232667212800, 113901166310400, 2193425619840000, 44398776748032000, 942498015750144000, 20938290999865344000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 2 of the general recurrence a(1) = 1, a(2) = 2*m+3, a(n+2) = (2*m+3)*a(n+1)+(n+1)*(n+3)*a(n), which arises when accelerating the convergence of a certain series for the constant log(2). For remarks on the general case see A142988 (m=0). For other cases see A142989 (m=1) and A142991 (m=3).

Seiichi Manyama, Table of n, a(n) for n = 1..446

Cf. A142979, A142983, A142988, A142989, A142991.

p := n -> (n^2-n+1)/3: a := n -> (n+2)!*p(n+2)*sum ((-1)^(k+1)/(k*(k+1)*(k+2)*p(k+1)*p(k+2)), k = 1..n): seq(a(n), n = 1..20);

a(n) = (n+2)!*p(n+2)*sum {k = 1..n} (-1)^(k+1)/(k*(k+1)*(k+2)*p(k+1)*p(k+2)), where p(n) = (n^2-n+1)/3. Recurrence: a(1) = 1, a(2) = 7, a(n+2) = 7*a(n+1)+(n+1)*(n+3)*a(n). The sequence b(n) := 1/2*(n+2)!*p(n+2) satisfies the same recurrence with b(1) = 7, b(2) = 52. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(7+1*3/(7+2*4/(7+3*5/(7+...+(n-1)*(n+1)/7)))), for n >=2. Lim n -> infinity a(n)/b(n) = 1/(7+1*3/(7+2*4/(7+3*5/(7+...+(n-1)*(n+1)/(7+...))))) = 2*sum {k = 1..inf} (-1)^(k+1)/ (k*(k+1)*(k+2)*p(k+1)*p(k+2)) = 24*log(2)-33/2.

A142991 a(1) = 1, a(2) = 9, a(n+2) = 9a(n+1)+(n+1)(n+3)*a(n).

Original entry on oeis.org

1, 9, 89, 936, 10560, 127800, 1657080, 22965120, 339252480, 5326819200, 88651670400, 1559600179200, 28929882240000, 564490975104000, 11560712397696000, 247991610230784000, 5561409662613504000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 3 of the general recurrence a(1) = 1, a(2) = 2*m+3, a(n+2) = (2*m+3)*a(n+1) + (n+1)*(n+3)*a(n), which arises when accelerating the convergence of a certain series for the constant log(2). For remarks on the general case see A142988 (m=0). For other cases see A142989 (m=1) and A142990 (m=2).

Seiichi Manyama, Table of n, a(n) for n = 1..445

Cf. A142979, A142983, A142988, A142989, A142990.

p := n -> (2*n^3-3*n^2+7*n-3)/15: a := n -> (n+2)!*p(n+2)*sum ((-1)^(k+1)/(k*(k+1)*(k+2)*p(k+1)*p(k+2)), k = 1..n): seq(a(n), n = 1..20);

RecurrenceTable[{a[1]==1,a[2]==9,a[n+2]==9a[n+1]+(n+1)(n+3)a[n]},a,{n,20}] (* Harvey P. Dale, Jul 18 2020 *)

a(n) = (n+2)!*p(n+2)*sum {k = 1..n} (-1)^(k+1)/(k*(k+1)*(k+2)*p(k+1)*p(k+2)), where p(n) = (2*n^3-3*n^2+7*n-3)/15. Recurrence: a(1) = 1, a(2) = 9, a(n+2) = 9*a(n+1)+(n+1)*(n+3)*a(n). The sequence b(n) := 1/2*(n+2)!*p(n+2) satisfies the same recurrence with b(1) = 9, b(2) = 84. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(9+1*3/(9+2*4/(9+3*5/(9+...+(n-1)*(n+1)/9)))), for n >=2. Lim n -> infinity a(n)/b(n) = 1/(9+1*3/(9+2*4/(9+3*5/(9+...+(n-1)*(n+1)/(9+...))))) = 2*sum {k = 1..inf} (-1)^(k+1)/ (k*(k+1)*(k+2)*p(k+1)*p(k+2)) = 167/6 - 40*log(2).

cp's OEIS Frontend

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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A142989 a(1) = 1, a(2) = 5, a(n+2) = 5*a(n+1)+(n+1)*(n+3)*a(n).

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A142990 a(1) = 1, a(2) = 7, a(n+2) = 7*a(n+1)+(n+1)*(n+3)*a(n).

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A142991 a(1) = 1, a(2) = 9, a(n+2) = 9*a(n+1)+(n+1)*(n+3)*a(n).

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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).

A142989 a(1) = 1, a(2) = 5, a(n+2) = 5a(n+1)+(n+1)(n+3)*a(n).

A142990 a(1) = 1, a(2) = 7, a(n+2) = 7a(n+1)+(n+1)(n+3)*a(n).

A142991 a(1) = 1, a(2) = 9, a(n+2) = 9a(n+1)+(n+1)(n+3)*a(n).