Synodic Cyclers
The EarthMars Synodic orbit is 780 days (2.135 years).
If a cycler period is exactly 2.135 Earth years, a given EarthMars angular relationship would repeat once per cycler orbit. We're particularly interested in the "correct" angular relationship where Earth's angular position leads position of Mars such that a cycler could pass near Earth then intersect orbit of Mars precisely when Mars gets there. This fortuitous happenstance would enable a payload of passengers and cargo to board cycler near Earth; then, depart cycler near Mars.
For this fortuitous double rendezvous to occur, the cycler's perihelion must itself lead this EarthMars angular relationship at certain angular interval. Thus, an optimum "shuttle" from Earth to Mars requires a simultaneous angular relation amongst all three celestial objects: 1) cycler perihelion must lead Earth by a certain angle; 2) Earth position must lead Mars by a certain angle. This circumstance would be a rare occurence.
With human help, this three part angular relationship could recur more frequently.
Describe Synodic Cycle
 
 
Typical reference ray (ν = 0°) extends from Sol to the perehelion, orbit's closest point to Sol. An asteroid is fastest at its perihelion; for this orbit, object takes only 50 days to travel the first 90° as shown. 
Asteroid is slowest at its farthest point (aphelion).
It takes 340 days from 90° point (Semilatus rectum) to 180° (aphelion).  This orbit is engineered for the asteriod ("cycler") to intersect Earth's orbit at ν=90°. For cycler to rendezvous with Earth, Terra's position 50 days prior to the rendezvous must lead the cycler by 41° as shown above.  After leaving Earth, cycler will subsequently intersect orbit of Mars at 112 days past perihelion postion. For cycler to rendezvous with Mars, the red planet must lead the cycler by 69° when cycler is at perihelion.  
First Year
 
Arbitrarily start EM synodic orbit such that Mars leads Earth by 28°. This same EM relation should reoccur at the completion of the Synodic Oribit.  After 50 days, cycler will rendezvous with Earth; pax and payload can board.
Lead of Mars reduces to only 5°.
 After only 62 more days, cycler can perform rendezvous with Mars.
Pax and payload can disembark.
 At the end of first year, cycler approaches the aphelion.  
Second Year and Beyond
 
After considerable progress around its orbit, cycler once again intercepts orbit of Mars, but Mars is literally nowhere to be seen.
It hides behind Sol!
 After 2 years, cycler once again intercepts orbit of Earth.
However, Earth is far away, and Mars still hides.
 Cycler completes its orbit, and the EarthMars angular relation repeats.
Mars again leads Earth by 28°.
 However, cycler can not repeat rendezvous with either Earth or Mars. They both lead cycler perihelion by 48.6° too far. 

Future content:
Multiple cyclers
inorbit fuel burns to adjust orbit
retrograde rendezvous, payloads from Mars to Earth
Synodic Cyclers. For cyclers with periods exactly 2.135 Earth years, their period would coincide with the Mars's synodic orbit viewed from Earth. Thus, when synodic cycler intersects Earth's orbit, the EarthMars angular relationship would be the same as the previous such intersection.
PROBLEM: This intersection point is not resonant and strays all over Earth's orbit. Thus, it would seldom be near Earth's position.
Earth (ⴲ) Orbit  Habitat (H) Orbit  Mars (♂) Orbit  

Aphelion  Q_{ⴲ} =  1.0 AU  Observed  Q_{H} =  2.5530 AU  = a + c  Q_{♂} =  1.6659 AU  Observed 
Perihelion  q_{ⴲ} =  1.0 AU  Observed  q_{H} =  0.6218 AU  = a  c  q_{♂} =  1.3815 AU  Observed 
Semimajor axis  a_{ⴲ} =  1.0 AU  = (Q + q) ÷ 2  a_{H} =  1.5874 AU  = ∛(T^{2})  a_{♂} =  1.5237 AU  = (Q + q) ÷ 2 
Semiminor axis  b_{ⴲ} =  1.0 AU  = (Q  q) ÷ 2  b_{H} =  1.2599 AU  = √(ℓ × a)  b_{♂} =  1.5170 AU  = (Q  q) ÷ 2 
Focus  c_{ⴲ} =  0.0 AU  = √(a^{2}  b^{2})  c_{H} =  0.9656 AU  = √(a^{2}  b^{2})  c_{♂} =  0.1422 AU  = √(a^{2}  b^{2}) 
Eccentricity  e_{ⴲ} =  0.0 AU  = c ÷ a  e_{H} =  0.6053  = c ÷ a  e_{♂} =  0.0933  = c ÷ a 
Semilatus Rectum  ℓ_{ⴲ} =  1.0 AU  = b^{2 }÷ a  ℓ_{H} =  1.00 AU  Given  ℓ_{♂} =  1.5104 AU  = b^{2 }÷ a 
Sidereal Period  T_{ⴲ} =  1.00 Yr  = √(a^{3})  T_{H} =  2.00 Yr  Given  T_{♂} =  1.88 Yr  = √(a^{3}) 
Angular Velocity  ω_{ⴲ} =  0.986°/dy  = 360°^{ }÷ T  ω_{H} =  Variable  TBD  ω_{♂} =  0.524°/dy  = 360° ÷ T 

Pos.  Time  Range  Cycler Velocity 

ν

Days
 AU 
km/sec

0°  0  0.614 AU  48.6 
90°  50  1.000 AU  35.2 
127°  112  1.611 AU  23.8 
180°  392  2.703 AU  11.02 
Given

Deg
 AU 
km/sec

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