OFFSET
1,2
COMMENTS
a(n) satisfies the recurrence a(n) = (2n-1)*a(n-1) - (n-1)*(n-2)*a(n-2) for n > 2, with initial conditions a(1)=1, a(2)=2.
The same recurrence is satisfied by A000262(n), but with different initial conditions.
The limit of a(n)/A000262(n) as n tends to infinity is the Euler-Gompertz constant G = e*E1(1), where E1 is an exponential integral. The decimal representation of G is given by A073003.
The convergents of the c.f. G = 1-1/(3-1*2/(5-2*3/(7-3*4/(9-...)))) are (a(n)/A000262(n)) = (1, 2/3, 8/13, 44/73, ...). The c.f. is equivalent to Bala's c.f. for 1-G given in the entry for A073003.
a(n)/A000262(n) - G ~ 2*Pi*exp(1-4*sqrt(n)) as n tends to infinity.
a(n)/n! ~ G*exp(2*sqrt(n))/(2*n^(3/4)*sqrt(Pi*e)) as n tends to infinity.
LINKS
Alois P. Heinz, Table of n, a(n) for n = 1..444
Richard P. Brent, M. L. Glasser, and Anthony J. Guttmann, A Conjectured Integer Sequence Arising From the Exponential Integral, arXiv:1812.00316 [math.NT], 2018.
NIST Digital Library of Mathematical Functions, Exponential, Logarithmic, Sine and Cosine Integrals.
FORMULA
a(n) = (2n-1)*a(n-1) - (n-1)*(n-2)*a(n-2) for n > 2.
E.g.f.: exp(x/(1-x))*(G - E1(x/(1-x))), where G is the Euler-Gompertz constant and E1 is an exponential integral.
Conjecture: Integral_{x = 0..oo} (x/(1 + x))^n*exp(-x) dx = 1/(n-1)!*( a(n) - A000262(n)*G ), where G = Integral_{x = 0..oo} exp(-x)/(1 + x) dx is the Euler-Gompertz constant A073003. - Peter Bala, Mar 20 2022
EXAMPLE
MAPLE
a:= proc(n) option remember; `if`(n<3, n,
(2*n-1)*a(n-1) -(n-1)*(n-2)*a(n-2))
end:
seq(a(n), n=1..23); # Alois P. Heinz, Dec 12 2018
MATHEMATICA
a[n_] := a[n] = (2n-1)a[n-1] - (n-1)(n-2)a[n-2]; a[1] = 1; a[2] = 2;
Array[a, 21] (* Jean-François Alcover, Oct 06 2019 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Richard P. Brent, Dec 12 2018
STATUS
approved