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.

A142988 a(1) = 1, a(2) = 3, a(n+2) = 3*a(n+1) + (n + 1)*(n + 3)*a(n).

Original entry on oeis.org

1, 3, 17, 96, 696, 5448, 49752, 492480, 5457600, 65128320, 850296960, 11864240640, 178442611200, 2848854758400, 48517709184000, 872011090944000, 16589133517824000, 331426982928384000, 6966369015201792000
Offset: 1

Views

Author

Peter Bala, Jul 17 2008

Keywords

Comments

This is the case m = 0 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) (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*(k + 1)*(k + 2)) = 2*(log(2) - 5/8). For other cases see A142989 (m = 1), A142990 (m = 2) and A142991 (m = 3).
The solution to the general recurrence may be expressed as a series: a(n) = (n+2)!*p_m(n+2)*Sum_{k = 1..n} (-1)^(k+1)/(k*(k + 1)*(k + 2)*p_m(k+1)*p_m(k+2)), where p_m(x) = 1/(2*x*(x - 1))*Sum_{k = 2..m+2} 2^k*C(m,k-2)*C(x,k).
The first few values are p_0(x) = 1, p_1(x) = (2*x - 1)/3, p_2(x) = (x^2 - x + 1)/3 and p_3(x) = (2*x^3 - 3*x^2 + 7*x - 3)/15.
The polynomial p_m(x) is the unique polynomial solution (up to multiplication by a constant) of the difference equation (x + 1)*f(x+1) - (x - 2)*f(x-1) = (2*m + 3)*f(x).
O.g.f. for the p_m(x): Sum_{k >= 0} p_m(x)*t^m = 1/(2*x*(x - 1)*t^2) * ((1 + t)^x/(1 - t)^(x-1) - 1 + t - 2*x*t) = 1 + (2*x - 1)/3*t + (x^2 - x + 1)/3*t^2 + ....
These polynomials satisfy a Riemann hypothesis: their zeros all lie on the vertical line Re x = 1/2 in the complex plane (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) = (1/2)*(n + 2)!*p_m(n+2), with b(1) = 2*m + 3, b(2) = 4*(m^2 + 3*m + 3).
Hence the behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = 2*Sum_{k >= 1} (-1)^(k+1)/(k*(k + 1)*(k + 2)*p_m(k+1)*p_m(k+2)) = 1/((2*m + 3) + 1*3/((2*m + 3) + 2*4/((2*m + 3) + 3*5/((2*m + 3) +...+ (n - 1)*(n + 1)/((2*m + 3) + ...))))).
The infinite continued fraction has the value (-1)^m*2*(m + 1)*(m + 2)*(log(2) - 5/8 - (1/2)*(1/(1*2*3) - 1/(2*3*4) + 1/(3*4*5) - ... + (-1)^(m+1)/(m*(m + 1)*(m + 2)))) for m > 0. This evaluation follows from a result of Ramanujan; see [Berndt, Chapter 12, Entry 34] (set l = n in Entry 34 and then let n tend to 1).
For related results see A142979 and A142983.

References

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

Links

Crossrefs

Cf. A142979, A142983, A142989, A142990, A142991.

Programs

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

Formula

a(n) = (n + 2)!*Sum_{k = 1..n} (-1)^(k+1)/(k*(k + 1)*(k + 2)).
Recurrence: a(1) = 1, a(2) = 3, a(n+2) = 3*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) = 3, b(2) = 12.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(3 + 1*3/(3 + 2*4/(3 + 3*5/(3 + ... + (n - 1)*(n + 1)/3)))), for n >= 2.
Limit_{n -> oo} a(n)/b(n) = 1/(3 + 1*3/(3 + 2*4/(3 + 3*5/(3 + ... + (n - 1)*(n + 1)/(3 + ...))))) = 2*Sum_{k >= 1} (-1)^(k+1)/(k*(k + 1)*(k + 2)) = 4*log(2) - 5/2.

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

Views

Author

Peter Bala, Jul 17 2008

Keywords

Comments

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).

Links

Crossrefs

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

Programs

  • Maple
    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);
  • Mathematica
    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 *)

Formula

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) = 7*a(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

Views

Author

Peter Bala, Jul 17 2008

Keywords

Comments

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).

Links

Crossrefs

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

Programs

  • Maple
    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);

Formula

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.
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.