%I #33 Mar 07 2024 15:49:23
%S 1,2,10,44,288,1896,15888,137952,1419840,15255360,186693120,
%T 2387093760,33898314240,502247692800,8123141376000,136785729024000,
%U 2483065912320000,46822564905984000,942853671825408000,19678282007924736000,435355106182520832000
%N a(1) = 1, a(2) = 2, a(n+2) = 2*a(n+1) + (n + 1)*(n + 2)*a(n).
%C This is the case m = 1 of the general recurrence a(1) = 1, a(2) = 2*m, a(n+2) = 2*m*a(n+1) + (n + 1)*(n + 2)*a(n) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series 1/2 + 1/2*Sum_{k > 1} (-1)^(k+1)/(k*(k + 1)) = log(2). For other cases see A142984 (m = 2), A142985 (m = 3), A142986 (m = 4) and A142987 (m = 5).
%C The solution to the general recurrence may be expressed as a sum: a(n) = n!*p_m(n+1)*Sum_{k = 1..n} (-1)^(k+1)/(p_m(k)*p_m(k+1)), where p_m(x) = Sum_{k = 1..m} 2^(k-1)*C(m-1,k-1)*C(x,k) is the polynomial that gives the regular polytope numbers for the m-dimensional cross polytope as defined by [Kim] (see A142978). The first few values are p_1(x) = x, p_2(x) = x^2, p_3(x) = (2*x^3 + x)/3 and p_4(x) = (x^4 + 2*x^2)/3.
%C The polynomial p_m(x) is the unique polynomial solution of the difference equation x*(f(x+1) - f(x-1)) = 2*m*f(x), normalized so that f(1) = 1.
%C The o.g.f. for the p_m(x) is 1/2*((1 + t)/(1 - t))^x = 1/2 + x*t + x^2*t^2 + (2*x^3 + x)/3*t^3 + .... Thus p_m(x) is, apart from a constant factor, the Meixner polynomial of the first kind M_m(x;b,c) at b = 0, c = -1, also known as a Mittag-Leffler polynomial.
%C The general recurrence in the first paragraph above has a second solution b(n) = n!*p_m(n+1) with b(1) = 2*m, b(2) = m^2 + 2. Hence the behavior of a(n) for large n is given by Limit_{n-> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(p_m(k)*p_m(k+1)) = 1/((2*m) + 1*2/((2*m) + 2*3/((2*m) + 3*4/((2*m) + ... + n*(n + 1)/((2*m) + ...))))) = 1 + (-1)^(m+1) * (2*m)*(log(2) - (1 - 1/2 + 1/3 - ... + (-1)^(m+1)/m)), where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 32(i)]).
%C See A142979, A142988 and A142992 for similar results. For corresponding results for Napier's constant e, the constant zeta(2) and Apery's constant zeta(3) refer to A000522, A142995 and A143003, respectively.
%D Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.
%H Vincenzo Librandi, <a href="/A142983/b142983.txt">Table of n, a(n) for n = 1..200</a>
%H Hyun Kwang Kim, <a href="http://dx.doi.org/10.1090/S0002-9939-02-06710-2">On Regular Polytope Numbers</a>, Proc. Amer. Math. Soc., 131 (2002), 65-75.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/MeixnerPolynomialoftheFirstKind.html">Meixner polynomial of the first kind</a>.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Mittag-LefflerPolynomial.html">Mittag-Leffler polynomial</a>.
%F a(n) = n!*p(n+1)*Sum_{k = 1..n} (-1)^(k+1)/(p(k)*p(k+1)), where p(n) = n.
%F Recurrence: a(1) = 1, a(2) = 2, a(n+2) = 2*a(n+1) + (n + 1)*(n + 2)*a(n).
%F The sequence b(n) := n!*p(n+1) satisfies the same recurrence with b(1) = 2 and b(2) = 6.
%F Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(2 + 1*2/(2 + 2*3/(2 + 3*4/(2 + ... + (n - 1)*n/2)))), for n >= 2.
%F The behavior of a(n) for large n is given by Limit_{n -> oo} a(n)/b(n) = 1/(2 + 1*2/(2 + 2*3/(2 + 3*4/(2 + ... + n*(n+1)/(2 + ...))))) = Sum_{k >= 1} (-1)^(k+1)/(k*(k + 1)) = 2*log(2) - 1.
%F E.g.f.: (2*log(x+1)-x)/(x-1)^2. - _Vaclav Kotesovec_, Oct 21 2012
%p a := n -> (n+1)!*sum ((-1)^(k+1)/(k*(k+1)), k = 1..n): seq(a(n), n = 1..20);
%t Rest[CoefficientList[Series[(-x+2*Log[x+1])/(x-1)^2,{x,0,20}],x]*Range[0,20]!] (* _Vaclav Kotesovec_, Oct 21 2012 *)
%o (Haskell)
%o a142983 n = a142983_list !! (n-1)
%o a142983_list = 1 : 2 : zipWith (+)
%o (map (* 2) $ tail a142983_list)
%o (zipWith (*) (drop 2 a002378_list) a142983_list)
%o -- _Reinhard Zumkeller_, Jul 17 2015
%Y Cf. A000522, A142979, A142984, A142985, A142986, A142987, A142988, A142992.
%Y Cf. A002378.
%K easy,nonn
%O 1,2
%A _Peter Bala_, Jul 17 2008