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%I #34 Mar 25 2022 15:51:34
%S 1,2,8,44,300,2420,22460,235260,2741660,35152820,491459820,7436765660,
%T 121046445260,2108118579060,39104985755420,769549656815420,
%U 16009942093608060,351030466622487860,8089084984387984460,195421894806240545820,4938445392988428283820
%N A sequence related to the Euler-Gompertz constant.
%C 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.
%C The same recurrence is satisfied by A000262(n), but with different initial conditions.
%C 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.
%C 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.
%C a(n)/A000262(n) - G ~ 2*Pi*exp(1-4*sqrt(n)) as n tends to infinity.
%C a(n)/n! ~ G*exp(2*sqrt(n))/(2*n^(3/4)*sqrt(Pi*e)) as n tends to infinity.
%C a(n) = A000262(n) - |A201203(n-2)| for n >= 2.
%H Alois P. Heinz, <a href="/A321942/b321942.txt">Table of n, a(n) for n = 1..444</a>
%H Richard P. Brent, M. L. Glasser, and Anthony J. Guttmann, <a href="https://arxiv.org/abs/1812.00316">A Conjectured Integer Sequence Arising From the Exponential Integral</a>, arXiv:1812.00316 [math.NT], 2018.
%H NIST Digital Library of Mathematical Functions, <a href="http://dlmf.nist.gov/6.2.1">Exponential, Logarithmic, Sine and Cosine Integrals</a>.
%F a(n) = (2n-1)*a(n-1) - (n-1)*(n-2)*a(n-2) for n > 2.
%F 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.
%F 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
%e a(3) = (2*3-1)*a(2) - 2*1*a(1) = 5*2 - 2*1 = 8.
%e a(3) = A000262(3) - |A201203(1)| = 13 - |5| = 8.
%p a:= proc(n) option remember; `if`(n<3, n,
%p (2*n-1)*a(n-1) -(n-1)*(n-2)*a(n-2))
%p end:
%p seq(a(n), n=1..23); # _Alois P. Heinz_, Dec 12 2018
%t a[n_] := a[n] = (2n-1)a[n-1] - (n-1)(n-2)a[n-2]; a[1] = 1; a[2] = 2;
%t Array[a, 21] (* _Jean-François Alcover_, Oct 06 2019 *)
%Y Cf. A000262, A073003, A201203.
%K nonn,easy
%O 1,2
%A _Richard P. Brent_, Dec 12 2018
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