Do you have the datasheet for your specific 4017? It is quite possible that you need more current than the 4017 can handle.
For example, the datasheet for a Texas Instruments CD4017BE shows (graphs on page 3) that with a drain-to-source voltage of 10V, each output pin can sink no more than ~15mA:
As @Tony mentions in his answer, if the LEDs have a typical forward voltage of 2 to 2.2 volts, then the 100Ω current limiting resistors you are using drop approximately 7 volts. Using Ohm's law, I = E / R, the current each output pin would need to sink is around 70mA.
Try a higher resistor value to lower the current significantly, say 10mA. The LEDs won't be as bright, but it might be within your 4017's current sinking ability and at least allow you to determine if things are working correctly.
You can figure out the resistor value to use by rearranging Ohm's law:
$$R = \frac{E}{I} = \frac{7}{0.02} = 350\Omega$$
If that works, and you want to make them brighter, you'll need to either find a 4017 that has greater current ratings on the output pins or add some transistors.
Edit:
Note that the circuit you linked has a "1K2" resistor connecting the LEDs to ground (negative). That's equivalent to 1.2kΩ which works out to slightly less than 6mA per LED.
Edit 2:
Per @Curd's comment, I am also including the source current diagram from the datasheet.
I think if you use a second 555, also connected as a monostable, but replace the timing capacitor with a short (and you don't need the timing resistor)-- triggered by the first one, that should work.
You could also use a D flip-flop such as a 4013 with the D tied high, but you'd need an inverter on the clock and a power-on reset circuit.
Or, something like this:
simulate this circuit – Schematic created using CircuitLab
C1/R3/D1 provides a semi-reliable power-on reset (100% reliable if the power turns on cleanly every time after a relatively long 'off' time where +V falls to near 0V).
C2 and R4 provide an edge trigger from the 555 output falling edge.
R1 and R2 protect the CMOS inputs
Best Answer
A 555 is about 1/10 the price, but accuracy will likely be less. It also has a plethora of 2nd sources.
The LTC6994 may be lower power, and has most likely higher accuracy.
Of course you could use a microcontroller and do better than either of the above in everything but cost, where the 555 still wins.