A242091 a(n) = r * (n-1)! where r is the rational number that satisfies the equation Sum_{k>=n} (-1)^(k + n)/C(k,n) = n*2^(n-1)*log(2) - r.

0, 2, 15, 128, 1310, 15864, 222936, 3572736, 64354608, 1287495360, 28328889600, 679936896000, 17678878214400, 495015296025600, 14850552286080000, 475219068007219200, 16157470542709708800, 581669316147767500800, 22103440771676298854400

Offset: 1

Author

Richard R. Forberg, Aug 14 2014

Keywords

• nonn

Comments

The sum of the terms of the inverse of the binomial coefficients, 1/C(k,n), with alternating signs, equals an irrational number which is expressed as m * log(2) - r, where m is the integer n*2^(n-1) = A001787(n), n>=1, and r is rational. a(n) = r * (n-1)!.

Examples

			Sum_{k>=1} (-1)^(k + 1)/C(k,1) = Sum_{k>=1} (-1)^(k + 1)/k = log(2) where m = 1 and r = 0. (See A002162.)
Sum_{k>=2} (-1)^(k + 2)/C(k,2) = 4*log(2) - 2. (See A000217.)
Sum_{k>=3} (-1)^(k + 3)/C(k,3) = 12*log(2) - 15/2. (See A000292.)
Sum_{k>=4} (-1)^(k + 4)/C(k,4) = 30*log(2) - 64/3. (See A000332.)
Sum_{k>=5} (-1)^(k + 5)/C(k,5) = 80*log(2) - 655/12. (See A000389.)
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..400

Crossrefs

Cf. A007318, A242023, A242024, A000217, A000292, A000332, A000389, A000579, A000580, etc. and A002162, A001787, A068102.

Programs

Magma

[n le 1 select 0 else 2*(n)*Self(n-1)+(Factorial(n) div (n-1)): n in [1..20]]; // Vincenzo Librandi, Sep 22 2015

Maple

seq(add(2^(n-j-1)*n!/j, j=1..n-1), n=1..100); # Robert Israel, Aug 14 2014

Mathematica

Table[Sum[2^(n - j - 1)*n!/j, {j, n - 1}], {n, 20}] (* Wesley Ivan Hurt, Aug 14 2014 *)
FullSimplify[Table[-1/2*n!*(LerchPhi[1/2, 1, n] - 2^n*Log[2]),{n, 1, 20}]] (* Vaclav Kotesovec, Aug 15 2014 *)

PARI

my(x='x+O('x^30)); concat([0], Vec(serlaplace(x*log(1-x)/(2*x-1)))) \\ G. C. Greubel, Nov 25 2017

Formula

From Robert Israel, Aug 14 2014: (Start)

a(n) = n * A068102(n-1).

a(n) = n! * Sum_{j=1..(n-1)} 2^(n-j-1)/j.

a(n) = n! * (2^(n-1)*log(2)-(1/2)*LerchPhi(1/2, 1, n)).

a(n+1) = 2*(n+1)*a(n) + (n+1)!/n.

E.g.f.: x*log(1-x)/(2*x-1).

(End)

Recurrence: (n-1)*a(n) = n*(3*n-4)*a(n-1) - 2*(n-2)*(n-1)*n*a(n-2). - Vaclav Kotesovec, Aug 15 2014