A142980 -id:A142980 - 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)

A135148 A binomial recursion: a(n) = q(n) (see formula).

Original entry on oeis.org

0, 1, 6, 45, 400, 4115, 48146, 631729, 9189972, 146829039, 2556200086, 48167698733, 976792093784, 21211601837803, 491112582793626, 12077021182230057, 314362864408454236, 8635229233659916007, 249631741661080132766, 7575921686807827601701, 240827454421807200901728

Offset: 1

Benoit Cloitre, Nov 20 2007

Vaclav Kotesovec, Table of n, a(n) for n = 1..400

Cf. A135147, A135149, A135150, A135074, A135075, A142980.

z[1] := x; z[n_] := 1 + Sum[(2 + Binomial[n, k])*z[k], {k, 1, n - 1}]; Table[ Coefficient[z[n], x, 0], {n, 1, 20}] (* G. C. Greubel, Sep 28 2016 *)
z[1] := x; z[n_] := z[n] = Expand[1 + Sum[(2 + Binomial[n, k])*z[k], {k, 1, n-1}]]; Table[Coefficient[z[n], x, 0], {n, 1, 30}] (* Vaclav Kotesovec, Nov 25 2020 *)
nmax = 30; Rest[CoefficientList[Series[(1 - E^x)*(E^x - 2*x - 1)/(2*(2 - E^x)^2), {x, 0, nmax}], x] * Range[0, nmax]!] (* Vaclav Kotesovec, Nov 25 2020 *)

r=1; s=2; v=vector(120, j, x); for(n=2, 120, g=r+sum(k=1, n-1, (s+binomial(n, k))*v[k]); v[n]=g); z(n)=v[n]; p(n)=polcoeff(z(n), 1); q(n)=polcoeff(z(n), 0); a(n)=p(n);

Let z(1) = x and z(n) = 1 + Sum_{k=1..n-1} (2 + binomial(n,k))*z(k), then z(n) = p(n)*x + q(n).
Limit_{n->oo} p(n)/q(n) = (3 - 2*log(2))/(2*log(2) - 1) = 4.177398899124179661610768...
a(n) ~ (2*log(2) - 1) * n * n! / (8 * log(2)^(n+2)). - Vaclav Kotesovec, Nov 25 2020
E.g.f.: (1 - exp(x)) * (exp(x) - 2*x - 1) / (2*(2 - exp(x))^2). - Vaclav Kotesovec, Nov 25 2020
a(n+1) = Sum_{k = 1..n} Stirling2(n, k)*A142980(k). - Peter Bala, Dec 10 2024

A142981 a(1) = 1, a(2) = 7, a(n+2) = 7a(n+1) + (n+1)^2a(n).

Original entry on oeis.org

1, 7, 53, 434, 3886, 38052, 406260, 4708368, 58959216, 794092320, 11454567840, 176267145600, 2883327788160, 49972442123520, 914939341344000, 17648374867200000, 357763095454464000, 7604722004802048000, 169148296960860672000, 3929342722459564032000, 95164717841561217024000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 3 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), which arises when accelerating the convergence of Mercator's series for the constant log(2). See A142979 for remarks on the general case.

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

Seiichi Manyama, Table of n, a(n) for n = 1..447

Cf. A001845, A024167, A142979, A142980, A142982.

p := n -> (4*n^3+6*n^2+8*n+3)/3: a := n -> n!*p(n)*sum ((-1)^(k+1)/(k*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 1..20)

Module[{a, n}, RecurrenceTable[{a[n+2] == 7*a[n+1] + (n+1)^2*a[n], a[1] == 1, a[2] == 7}, a, {n, 25}]] (* Paolo Xausa, Dec 12 2024 *)

a(n) = n!*p(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p(k-1)*p(k)), where p(n) = (4*n^3 + 6*n^2 + 8*n + 3)/3 = A001845(n) is the Ehrhart polynomial for the 3-dimensional cross polytope (the octahedron).
Recurrence: a(1) = 1, a(2) = 7, a(n+2) = 7*a(n+1) + (n + 1)^2*a(n). The sequence b(n):= n!*p(n) satisfies the same recurrence with b(1) = 7, b(2) = 50.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(7 + 1^2/(7 + 2^2/(7 + 3^2/(7 + ... + (n-1)^2/7)))), for n >= 2.
The behavior of a(n) for large n is given by limit_{n -> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(k*p(k-1)*p(k)) = 1/(7 + 1^2/(7 + 2^2/(7 + 3^2/(7 + ... + n^2/(7 + ...))))) = (1 - 1/2 + 1/3) - log(2); the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 29]). Thus a(n) ~ c*n^3*n! as n -> oo, where c = (10 - 12*log(2))/9.
E.g.f.: A(x) = (2*x*(4*x^2 + 3*x + 3) - 3*(x + 1)^3*log(1 + x) )/(3*(1 - x)^4) satisfies the differential equation 1 + (x + 7)*A(x) + (x^2 - 1)*A'(x) = 0 with A(0) = 0. - Peter Bala, Dec 09 2024

A142982 a(1) = 1, a(2) = 9, a(n+2) = 9a(n+1) + (n + 1)^2a(n).

Original entry on oeis.org

1, 9, 85, 846, 8974, 101916, 1240308, 16156656, 224789616, 3331795680, 52465122720, 875333381760, 15432978107520, 286828144485120, 5606317009440000, 114993185594112000, 2470155824763648000, 55464433059571200000, 1299510384759562752000, 31718253797341267968000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 4 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), which arises when accelerating the convergence of Mercator's series for the constant log(2). See A142979 for remarks on the general case.

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

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

Cf. A024167, A142979, A142980, A142981.

p := n -> (2*n^4+4*n^3+10*n^2+8*n+3)/3: a := n -> n!*p(n)*sum ((-1)^(k+1)/(k*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 1..20);

Module[{a, n}, RecurrenceTable[{a[n+2] == 9*a[n+1] + (n+1)^2*a[n], a[1] == 1, a[2] == 9}, a, {n, 25}]] (* Paolo Xausa, Dec 12 2024 *)

a(n) = n!*p(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p(k-1)*p(k)), where p(n) = (2*n^4 + 4*n^3 + 10*n^2 + 8*n + 3)/3 = A001846(n) is the Ehrhart polynomial for the 4-dimensional cross polytope (the 16-cell).
Recurrence: a(1) = 1, a(2) = 9, a(n+2) = 9*a(n+1) + (n + 1)^2*a(n).
The sequence b(n) := n!*p(n) satisfies the same recurrence with b(1) = 9, b(2) = 82.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(9 + 1^2/(9 + 2^2/(9 + 3^2/(9 + ... + (n-1)^2/9)))), for n >= 2.
The behavior of a(n) for large n is given by limit_{n -> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(k*p(k-1)*p(k)) = 1/(9 + 1^2/(9 + 2^2/(9 + 3^2/(9 + ... + n^2/(9 + ...))))) = log(2) - (1 - 1/2 + 1/3 - 1/4); the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 29]).
Thus a(n) ~ c*n^4*n! as n -> oo, where c = (12*log(2) - 7)/18.
E.g.f.: A(x) = (3*(x + 1)^4*log(1 + x) - 4*x^2*(2*x^2 + 2*x + 3))/(3*(1 - x)^5) satisfies the differential equation 1 + (x + 9)*A(x) + (x^2 - 1)*A'(x) = 0 with A(0) = 0. - Peter Bala, Dec 09 2024

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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A135148 A binomial recursion: a(n) = q(n) (see formula).

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A142981 a(1) = 1, a(2) = 7, a(n+2) = 7*a(n+1) + (n+1)^2*a(n).

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A142982 a(1) = 1, a(2) = 9, a(n+2) = 9*a(n+1) + (n + 1)^2*a(n).

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

A142981 a(1) = 1, a(2) = 7, a(n+2) = 7a(n+1) + (n+1)^2a(n).

A142982 a(1) = 1, a(2) = 9, a(n+2) = 9a(n+1) + (n + 1)^2a(n).