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

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

Offset

0,3

Comments

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]

Links

G. C. Greubel, Table of n, a(n) for n = 0..390

K. S. Brown, Integer Sequences Related To Pi

Formula

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

Maple

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

Mathematica

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 *)

Crossrefs

Cf. A155729, A012244, A054766.

Sequence in context: A307697 A320352 A301921 * A362660 A232691 A057719

Adjacent sequences: A054762 A054763 A054764 * A054766 A054767 A054768

Keyword

nonn,easy

Author

N. J. A. Sloane, May 26 2000

Extensions

More terms from James A. Sellers, May 27 2000

Status

approved