What is this opamp configuration?
In answer to the question posed by the title, this is a transimpedance amplifier with a resistor across the input to help set the DC level. 


The (ideal) gain of the transimpedance amplifier is Vo/Iin = 36K (just the value of R2). So peak-to-peak output from a 1uA peak input should be 72mV (ignoring the roll-off).  R1 has virtually no effect on the gain, it just makes the output noise a bit worse. 
The voltage change seen at the inverting input is as a result of the finite gain of the amplifier, ideally it would be zero. It should be the output voltage change divided by the open-loop gain of the amplifier. 
Your formula looks correct. 


If you wish to have the output voltage (at zero input current) equal the 1.65V reference in your top circuit, just lose the 47K resistor. There will be a DC 1.65V across your input current source. Of course you perhaps could connect the low side of the current source to Vref rather than ground. Or put a large capacitor in series with the input for AC coupling.