

A054765


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


6



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


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*n1)*procname(n1)+(n1)^2*procname(n2) ;
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: A232607 A320352 A301921 * A232691 A057719 A289258
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



