A321942 A sequence related to the Euler-Gompertz constant.

1, 2, 8, 44, 300, 2420, 22460, 235260, 2741660, 35152820, 491459820, 7436765660, 121046445260, 2108118579060, 39104985755420, 769549656815420, 16009942093608060, 351030466622487860, 8089084984387984460, 195421894806240545820, 4938445392988428283820

Offset: 1

Author

Richard P. Brent, Dec 12 2018

Keywords

• nonn
• easy

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.

a(n) = A000262(n) - |A201203(n-2)| for n >= 2.

Examples

			a(3) = (2*3-1)*a(2) - 2*1*a(1) = 5*2 - 2*1 = 8.
a(3) = A000262(3) - |A201203(1)| = 13 - |5| = 8.
		

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.

Crossrefs

Cf. A000262, A073003, A201203.

Programs

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 *)

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