A142985 -id:A142985 - cp's OEIS frontend

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

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

A142984 a(1) = 1, a(2) = 4, a(n+2) = 4a(n+1) + (n + 1)(n + 2)*a(n).

Original entry on oeis.org

1, 4, 22, 136, 984, 8016, 73392, 742464, 8254080, 99838080, 1307301120, 18407831040, 277570298880, 4460506444800, 76131788544000, 1375048700928000, 26208041287680000, 525597067634688000, 11065538390925312000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 2 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), which arises when accelerating the convergence of a certain series for the constant log(2). See A142983 for remarks on the general case.

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

Reinhard Zumkeller, Table of n, a(n) for n = 1..250

Cf. A142983, A142985, A142986, A142987.

Cf. A002378.

a142984 n = a142984_list !! (n-1)
a142984_list = 1 : 4 : zipWith (+)
                       (map (* 4) $ tail a142984_list)
                       (zipWith (*) (drop 2 a002378_list) a142984_list)
-- Reinhard Zumkeller, Jul 17 2015

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

a(n) = n!*p(n+1)*Sum_{k = 1..n} (-1)^(k+1)/(p(k)*p(k+1)), where p(n) = n^2.
Recurrence: a(1) = 1, a(2) = 4, a(n+2) = 4*a(n+1) + (n + 1)*(n + 2)*a(n).
The sequence b(n) := n!*p(n+1) satisfies the same recurrence with b(1) = 4 and b(2) = 18.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(4 + 1*2/(4 + 2*3/(4 + 3*4/(4 + ... + n*(n - 1)/(4))))), for n >= 2.
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*(k + 1)^2) = 1/(4 + 1*2/(4 + 2*3/(4 + 3*4/(4 + ... + n*(n - 1)/(4 + ...))))) = 3 - 4*log(2), where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 32(i)]).
a(n) = n! * (((-1)^n * (2 * Psi(n/2 + 1) - 2 * Psi(n/2 + 3/2)) - 4 * log(2) + 3) * (n+1)^2 + (-1)^n * (2 * n + 1)). - Robert Israel, Mar 07 2024

A142986 a(1) = 1, a(2) = 8, a(n+2) = 8a(n+1) + (n + 1)(n + 2)*a(n).

Original entry on oeis.org

1, 8, 70, 656, 6648, 72864, 862128, 10977408, 149892480, 2187106560, 33985025280, 560578268160, 9786290088960, 180315565516800, 3497645442816000, 71256899266560000, 1521414754578432000, 33975929212194816000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 4 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), which arises when accelerating the convergence of a certain series for the constant log(2). See A142983 for remarks on the general case.

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

Reinhard Zumkeller, Table of n, a(n) for n = 1..250

Cf. A014820, A142983, A142984, A142985, A142987.

Cf. A002378.

a142986 n = a142986_list !! (n-1)
a142986_list = 1 : 8 : zipWith (+)
                       (map (* 8) $ tail a142986_list)
                       (zipWith (*) (drop 2 a002378_list) a142986_list)
-- Reinhard Zumkeller, Jul 17 2015

p := n -> (n^4+2*n^2)/3: a := n -> n!*p(n+1)*sum ((-1)^(k+1)/(p(k)*p(k+1)), k = 1..n): seq(a(n), n = 1..20);

RecurrenceTable[{a[1]==1,a[2]==8,a[n]==8a[n-1]+n(n-1)a[n-2]},a,{n,20}] (* Harvey P. Dale, Apr 08 2015 *)

a(n) = n!*p(n+1)*Sum_{k = 1..n} (-1)^(k+1)/(p(k)*p(k+1)), where p(n) = (n^4 + 2*n^2)/3 = A014820(n).
Recurrence: a(1) = 1, a(2) = 8, a(n+2) = 8*a(n+1) + (n + 1)*(n + 2)*a(n).
The sequence b(n) := n!*p(n+1) satisfies the same recurrence with b(1) = 8 and b(2) = 66.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(8 + 1*2/(8 + 2*3/(8 + 3*4/(8 + ... + n*(n - 1)/8)))), for n >= 2.
The behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(p(k)*p(k+1)) = 1/(8 + 1*2/(8 + 2*3/(8 + 3*4/(8 + ... + n*(n - 1)/(8 + ...))))) = 17/3 - 8*log(2), where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 32(i)]).

A142987 a(1) = 1, a(2) = 10, a(n+2) = 10a(n+1) + (n + 1)(n + 2)*a(n).

Original entry on oeis.org

1, 10, 106, 1180, 13920, 174600, 2330640, 33084000, 498646080, 7964020800, 134491276800, 2396163513600, 44942274316800, 885524502643200, 18293122632960000, 395457106963968000, 8930300425804800000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 5 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), which arises when accelerating the convergence of a certain series for the constant log(2). See A142983 for remarks on the general case.

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

Reinhard Zumkeller, Table of n, a(n) for n = 1..250

Cf. A069038, A142983, A142984, A142985, A142986.

Cf. A002378.

a142987 n = a142987_list !! (n-1)
a142987_list = 1 : 10 : zipWith (+)
                        (map (* 10) $ tail a142987_list)
                        (zipWith (*) (drop 2 a002378_list) a142987_list)
-- Reinhard Zumkeller, Jul 17 2015

p := n -> (2*n^5+10*n^3+3*n)/15: a := n -> n!*p(n+1)*sum ((-1)^(k+1)/(p(k)*p(k+1)), k = 1..n): seq(a(n), n = 1..20);

RecurrenceTable[{a[1]==1,a[2]==10,a[n+2]==10a[n+1]+(n+1)(n+2)a[n]},a,{n,20}] (* Harvey P. Dale, Mar 23 2021 *)

a(n) = n!*p(n+1)*Sum_{k = 1..n} (-1)^(k+1)/(p(k)*p(k+1)), where p(n) = (2*n^5 + 10*n^3 + 3*n)/15 = A069038(n).
Recurrence: a(1) = 1, a(2) = 10, a(n+2) = 10*a(n+1) + (n + 1)*(n + 2)*a(n).
The sequence b(n) := n!*p(n+1) satisfies the same recurrence with b(1) = 10 and b(2) = 102.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(10 + 1*2/(10 + 2*3/(10 + 3*4/(10 + .. +n*(n - 1)/(10))))), for n >= 2.
The behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(p(k)*p(k+1)) = 1/(10 + 1*2/(10 + 2*3/(10 + 3*4/(10 + ... + n*(n - 1)/(10 + ...))))) = 10*log(2) - 41/6, where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 32(i)]).

cp's OEIS Frontend

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

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A142984 a(1) = 1, a(2) = 4, a(n+2) = 4*a(n+1) + (n + 1)*(n + 2)*a(n).

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A142986 a(1) = 1, a(2) = 8, a(n+2) = 8*a(n+1) + (n + 1)*(n + 2)*a(n).

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Author

Keywords

Comments

References

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

Formula

A142987 a(1) = 1, a(2) = 10, a(n+2) = 10*a(n+1) + (n + 1)*(n + 2)*a(n).

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

Formula

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

A142984 a(1) = 1, a(2) = 4, a(n+2) = 4a(n+1) + (n + 1)(n + 2)*a(n).

A142986 a(1) = 1, a(2) = 8, a(n+2) = 8a(n+1) + (n + 1)(n + 2)*a(n).

A142987 a(1) = 1, a(2) = 10, a(n+2) = 10a(n+1) + (n + 1)(n + 2)*a(n).