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.

Showing 1-3 of 3 results.

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

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Author

Peter Bala, Jul 17 2008

Keywords

Comments

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.

References

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

Links

Crossrefs

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

Programs

  • Maple
    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)
  • Mathematica
    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 *)

Formula

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)

A142980 a(1) = 1, a(2) = 5, a(n+2) = 5*a(n+1) + (n + 1)^2*a(n).

Original entry on oeis.org

1, 5, 29, 190, 1414, 11820, 110004, 1129200, 12686256, 154896480, 2043108000, 28958014080, 438997622400, 7088892491520, 121487996448000, 2202440792832000, 42113131054848000, 847071044402688000, 17880009683784192000, 395192695448291328000, 9127967350755133440000
Offset: 1

Views

Author

Peter Bala, Jul 17 2008

Keywords

Comments

This is the case m = 2 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.

Links

Crossrefs

Cf. A024167, A135148, A142979, A142981, A142982.

Programs

  • Maple
    p := n -> 2*n^2+ 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)
  • Mathematica
    Module[{a, n}, RecurrenceTable[{a[n+2] == 5*a[n+1] + (n+1)^2*a[n], a[1] == 1, a[2] == 5}, a, {n, 25}]] (* Paolo Xausa, Dec 12 2024 *)

Formula

a(n) = n!*p(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p(k-1)*p(k)), where p(n) = 2*n^2 + 2*n + 1 = A001844(n) is the Ehrhart polynomial for the 2-dimensional cross polytope (a square).
Recurrence: a(1) = 1, a(2) = 5, a(n+2) = 5*a(n+1) + (n+1)^2*a(n).
The sequence b(n) := n!*p(n) satisfies the same recurrence with b(1) = 5, b(2) = 26.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(5 + 1^2/(5 + 2^2/(5 + 3^2/(5 + ... + (n-1)^2/5)))), for n >= 2.
The behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = 1/(5 + 1^2/(5 + 2^2/(5 + 3^2/(5 + ... + n^2/(5 + ...))))) = Sum_{k >= 1} (-1)^(k+1)/(k*(4*k^4 + 1)) = log(2) - (1 - 1/2); the final equality is a result of Glaisher.
Thus a(n) ~ c*n^2*n! as n -> oo, where c = 2*log(2) - 1.
From Peter Bala, Dec 09 2024: (Start)
E.g.f.: A(x) = ((1 + x)^2 *log(1 + x) - 2*x^2)/(1 - x)^3 satisfies the differential equation 1 + (x + 5)*A(x) + (x^2 - 1)*A(x)' with A(0) = 0.
Sum_{k = 1..n} Stirling2(n, k)*a(k) = A135148(n+1). (End)

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

Views

Author

Peter Bala, Jul 17 2008

Keywords

Comments

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.

References

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

Links

Crossrefs

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

Programs

  • Maple
    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)
  • Mathematica
    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 *)

Formula

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
Showing 1-3 of 3 results.

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