It's often better to do nothing at all than to take two steps back an only one step forward

According to the Holographic principle the number of "bits" associated with a horizon (such as the event horizon of a black hole or Hubble horizon of the observable universe where objects recede from us at the speed of light due to the expansion of space and which also limits the patch that is causally connected to us in the cosmos) is proportional to the area of the horizon (in Planck units).

If one assumes the total energy of such patch of cosmos is distributed evenly among the horizon degrees of freedom according to the thermodynamic equipartition theorem, there exists a corresponding force as a result of the tendency of the universe to increase its entropy (second law of thermodynamics). This mechanism may be responsible for both the tradiational forms of gravity and the cosmological constant.

Already a fairly naive calculation finds that the value of the cosmological constant (associated with dark energy) would be in the correct ballpark respect to the observations. If this mechanism is real then there might be no need for any special dark energy.

T = 13.8e9; % current age of the universe (in years)
G = 6.674e-11;
c = 299792458;
hbar = 1.0546e-34;
ly = 9.4607e15; % lightyear (in meters)
lp = sqrt(hbar*G/c^3); % Planck length

R = T*ly % radius of causally connected universe (in meters) or c/H (Hubble horizon)
r = R/lp % radius in Planck lengths
N = pi*r^2 % number of bits associated with the Hubble horizon

  % btw. this value of N is approx. 2^407 which when interpreted (kind of) classically
  % suggests that no quantum computer with more than 407 ideal qubits will work as expected
  % but if these are qubits instead of bits then no such conclusion is true

  % Schwarzschild radius of the Hubble sphere: rs = 2*G*M/c^2 =>

M = R*c^2/2/G         % SI units
m = M/sqrt(hbar*c/G); % in Planck units

  % [arXiv:1001.0785v1]
  % equipartition of energy among the horizon degrees of freedom: 0.5*kT = m/(pi*r^2)

kT = m/(pi*r^2)/0.5

  % entropic force F = kT * gradient(N) =>
  
F = kT*2*pi*r;

  % Second Friedmann equation: a''/a = -4*pi*(rho/3 + p) + Lambda/3
  % Cosmic expansion: r''/r = F/(mr)

F/(m*r) % ~ 6e-122 so we are in the correct ballpark [arXiv:1105.3105v1]
4/r^2   % m is actually irrelevant if you model the universe this way, but 
        % you could also just take the directly observed values and arrive at similar lambda

  % additionally by combining the Unruh temperature T = a*hbar/(2*pi*c*k),
  % Newton's second law F = ma and area of holographic surface A = 4*pi*r^2,
  % one arrives at the Newton's law of universal gravitation F = GmM/r^2