A054766 -id:A054766 - cp's OEIS frontend

A012244 a(n+2) = (2n+3)a(n+1) + (n+1)^2a(n), a(0) = 1, a(1) = 1.

1, 1, 4, 24, 204, 2220, 29520, 463680, 8401680, 172504080, 3958113600, 100370793600, 2787459998400, 84139894238400, 2742857884166400, 96034297911552000, 3594206259195552000, 143193586818810528000, 6050501147565883008000, 270263264589232282368000

Offset: 0

N. J. A. Sloane

a(n) is the number of n-letter words from an n-letter alphabet such that no letter appears more than twice. - Paul Boddington, Nov 17 2003

Alois P. Heinz, Table of n, a(n) for n = 0..393
K. S. Brown, Integer Sequences Related To Pi
D. Dominici, Nested derivatives: A simple method for computing series expansions of inverse functions, arXiv:math/0501052 [math.CA], 2005.

Cf. A089975, A190392, A054765, A054766.

f := proc(n) option remember; if n <= 1 then 1 else (2*n-1)*f(n-1) +(n-1)^2*f(n-2); fi; end;

Range[0,20]! CoefficientList[Series[1/(1-2x-x^2)^(1/2), {x,0,20}], x]  (* Geoffrey Critzer, Dec 07 2011 *)

{a(n)=local(X=x+x^2*O(x^n));(n+1)!*polcoeff(serreverse(cos(X)+sin(X)-1),n+1)} \\ Paul D. Hanna, Aug 08 2012

E.g.f.: A(x) = (1 - 2*x - x^2)^(-1/2). - Paul Boddington, Nov 17 2003
a(n) = n!/2^n*A006139(n) = n!*Sum_{k=floor(n/2)..n} 2^(k-n)*C(n, k)*C(k, n-k). Sum_{n>=0} a(n)*x^n/n!^2 = exp(x)*BesselI(0, sqrt(2)*x). a(n) is the central coefficient of n!*(1+x+x^2/2)^n. - Vladeta Jovovic, Mar 22 2004
From Peter Bala, Aug 25 2011: (Start)
The function B(x) := int {t=0..x} A(t), obtained by integrating the generating function A(x), satisfies the autonomous differential equation d/dx(B(x)) = 1/(cos(B(x))-sin(B(x))). Compare with A190392.
Thus B(x), and hence A(x), can be found by inverting the function int {t=0..x} (cos(t)-sin(t)). By applying [Dominici, Theorem 4.1] the result can be expressed as
A(x) = 1 + sum {n>=1} D^n[1/(cos(t)-sin(t))](0)*x^n/n!, where the nested derivative D^n[f](x) of a function f(x) is defined recursively as D^0[f](x) = 1 and D^(n+1)[f](x) = d/dx(f(x)*D^n[f](x)) for n >= 0. Thus a(n) = D^n[1/(cos(t)-sin(t))](0). (End)
E.g.f. at offset 1: Series_Reversion(cos(x) + sin(x) - 1). - Paul D. Hanna, Aug 08 2012
a(n) ~ (1+sqrt(2))^(n+1/2) * n^n / (2^(1/4) * exp(n)). - Vaclav Kotesovec, Feb 18 2017

A054765 a(n+2) = (2n+3)a(n+1) + (n+1)^2a(n), a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 3, 19, 160, 1744, 23184, 364176, 6598656, 135484416, 3108695040, 78831037440, 2189265960960, 66083318415360, 2154235544616960, 75425161203302400, 2822882994841190400, 112463980097804697600

Offset: 0

N. J. A. Sloane, May 26 2000

The denominators of the convergents of [1/3, 4/5, 9/7, 16/9, ...]. To produce Pi the above continued fraction is used. It is formed by n^2/(2*n+1) which starts at n=1. Most numerators of continued fractions are 1 & thus are left out of the brackets. In the case of Pi they vary. Therefore here both numerators & denominators are given. The first 4 convergents are 1/3,5/19,44/160,476/1744. The value of this continued fraction is .273239... . 4*INV(1+.273239...) is Pi. - Al Hakanson (hawkuu(AT)gmail.com), Dec 01 2008

Starting with offset 1 = row sums of triangle A155729. [Gary W. Adamson & Alexander R. Povolotsky, Jan 25 2009]

G. C. Greubel, Table of n, a(n) for n = 0..390
K. S. Brown, Integer Sequences Related To Pi

Cf. A155729, A012244, A054766.

A054765 := proc(n)
    option remember;
    if n <= 1 then
        n;
    else
        (2*n-1)*procname(n-1)+(n-1)^2*procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Jul 13 2013

RecurrenceTable[{a[n + 2] == (2*n + 3)*a[n + 1] + (n + 1)^2*a[n],
a[0] == 0, a[1] == 1}, a, {n,0,50}] (* G. C. Greubel, Feb 18 2017 *)

a(n) ~ Pi * (1+sqrt(2))^(n + 1/2) * n^n / (2^(9/4) * exp(n)). - Vaclav Kotesovec, Feb 18 2017

More terms from James Sellers, May 27 2000

A345125 Numerator of 4/(1 + 1^2/(3 + 2^2/(5 + 3^2/(7 + ... + (n-1)^2/(2*n-1) )))).

Original entry on oeis.org

0, 4, 3, 19, 160, 1744, 644, 2529, 183296, 3763456, 4317632, 54743776, 1013549056, 30594128896, 35618973952, 10392576224, 3111643512832, 123968232030208, 48501417558016, 1083228572868608, 4080033616887808, 188557135970304, 3781715948011520

Offset: 0

Seiichi Manyama, Sep 16 2021

The limit of a(n)/A345259(n) is Pi.

			4/(1 + 1^2/(3 + 2^2/5)) = 19/6. So a(3) = 19.
0, 4, 3, 19/6, 160/51, 1744/555, 644/205, 2529/805, 183296/58345, ...

Seiichi Manyama, Table of n, a(n) for n = 0..1000
Frits Beukers, A rational approach to Pi, Nieuw Archief voor de Wiskunde, December 2000, pp. 372-379.

Cf. A012244, A054765, A054766, A345259 (denominator).

nmax = 25; Join[{0}, Table[4/(1 + ContinuedFractionK[j^2, (2*j + 1), {j, 1, k}]), {k, 0, nmax}] // Numerator] (* Vaclav Kotesovec, Sep 16 2021 *)

a(n) = my(x=0); forstep(i=n, 2, -1, x = (i-1)^2/((2*i-1)+x);); if (n, numerator(4/(1+x)), numerator(x)); \\ Michel Marcus, Sep 16 2021

a(n)/A345259(n) = 4 * A054765(n)/A012244(n).

A345259 Denominator of 4/(1 + 1^2/(3 + 2^2/(5 + 3^2/(7 + ... + (n-1)^2/(2*n-1) )))).

Original entry on oeis.org

1, 1, 1, 6, 51, 555, 205, 805, 58345, 1197945, 1374345, 17425485, 322622685, 9738413685, 11337871545, 3308059755, 990466892415, 39460313827935, 15438480702645, 344802363740835, 1298715036217599, 60019600489849, 1203757572990973

Offset: 0

Seiichi Manyama, Sep 16 2021

The limit of A345125(n)/a(n) is Pi.

			4/(1 + 1^2/(3 + 2^2/5)) = 19/6. So a(3) = 6.
0, 4, 3, 19/6, 160/51, 1744/555, 644/205, 2529/805, 183296/58345, ...

Seiichi Manyama, Table of n, a(n) for n = 0..1000
Frits Beukers, A rational approach to Pi, Nieuw Archief voor de Wiskunde, December 2000, pp. 372-379.

Cf. A012244, A054765, A054766, A345125 (numerator).

nmax = 25; Join[{1}, Table[4/(1 + ContinuedFractionK[j^2, (2*j + 1), {j, 1, k}]), {k, 0, nmax}] // Denominator] (* Vaclav Kotesovec, Sep 16 2021 *)

a(n) = my(x=0); forstep(i=n, 2, -1, x = (i-1)^2/((2*i-1)+x);); if (n, denominator(4/(1+x)), denominator(x)); \\ Michel Marcus, Sep 16 2021

A345125(n)/a(n) = 4 * A054765(n)/A012244(n).

cp's OEIS Frontend

A012244 a(n+2) = (2n+3)*a(n+1) + (n+1)^2*a(n), a(0) = 1, a(1) = 1.

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A054765 a(n+2) = (2n+3)*a(n+1) + (n+1)^2*a(n), a(0) = 0, a(1) = 1.

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A345125 Numerator of 4/(1 + 1^2/(3 + 2^2/(5 + 3^2/(7 + ... + (n-1)^2/(2*n-1) )))).

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A345259 Denominator of 4/(1 + 1^2/(3 + 2^2/(5 + 3^2/(7 + ... + (n-1)^2/(2*n-1) )))).

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Mathematica

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Formula

A012244 a(n+2) = (2n+3)a(n+1) + (n+1)^2a(n), a(0) = 1, a(1) = 1.

A054765 a(n+2) = (2n+3)a(n+1) + (n+1)^2a(n), a(0) = 0, a(1) = 1.