Hi Folks,
After an hiatus of a few days, I have succeeded in finding a remarkably simple solution to my problem and will post it here for reference, as there isn't much on FDNR filters for the amateur on the web. Lots for professional engineers but not a lot for hobbyists. My mistake was trying to be too clever and the equation I was struggling to find was for the D element of an FDNR filter, which it turns out you don't really need!
I will use a simple 3rd order Chebychev low pass filter as an example. In these circuits, R1 (10Meg) in the simulation is the element which the output is being measured across and is there for the analysis only, it should not appear on the pcb.
First, using software or tables of your choice (I used the Filter Design program by the late AADE) create a passive LC filter of the required order and configuration. Capacitors should have one end grounded. Design impedance is 1ohm and frequency 1Hz, see attachment 1Hz 1ohm 3rd order Chebychev LPF AADE.png and if you were to run a simulation it will work just fine. However, the impedance of one ohm, the frequency of one hertz plus the one farad capacitors is a problem and this circuit is not practical.
To convert this to FDNR form, the circuit shown in 3rd order Chebychev low pass 1Hz 1ohm FDNR circuit.png is used and it is necessary to scale the values. R3 and R4 can both be changed to 1k (the end result of scaling is shown in 3rd order Chebychev low pass 10Hz 10kohm FDNR circuit.png, see later). To minimise the capacitances required, decide on a scaling factor for the circuit impedance. Let's choose 10k. Divide the 1F capacitances in the 1Hz 1ohm FDNR circuit by 10,000 to get 100uF and divide by another ten to increase the frequency to 10Hz from 1Hz, so these caps are now 10uF. The resistors R6 and R7 represent the inductors L3 and L5 in the AADE screenshot, so an inductance of 0.1954H multiplied by 10,000 becomes a resistance of 1954ohms. Finally, resistor R2 represents the AADE capacitor C4, and its value of 0.1834F is also multiplied by 10,000 to give R2 = 1834ohms. This completed circuit is shown in 3rd order Chebychev low pass 10Hz 10kohm FDNR circuit.png In practice, the 1954ohm resistors R6 and R7 would be 2k and R2 would be 1k8, and there would be no significant difference in frequency response. The series input capacitor and shunt output capacitor (terminators) are equivalents to the original 1ohm input and output resistors, scaled to 10uF, and are essential for correct operation.
The 1Meg resistors connected across the capacitors are there to allow for the input bias and offset currents of the op amps. Without these, the outputs would tend to go to one or other of the power rails and the filter would not work. As their value is 100x the circuit impedance, their effect on performance is negligible.
For MicroCap users, the source file is attached, 3rd order Chebychev FDNR 10Hz 10kohm.txt, you will need to modify the txt extension to .cir to get it to run.
The frequency response is shown in 3rd order Chebychev low pass 10Hz 10kohm FDNR ac analysis.png and is commendably flat in the passband
The noise analysis results are shown in 3rd order Chebychev low pass 10Hz 10kohm FDNR noise analysis.png and evidently there is nothing to worry about regarding the noise performance. It makes very little difference what op amp you use here, a low bias and offset current JFET input op amp like TL081 may even be preferable to a low noise chip like NE5532, but the op amp must be unity gain stable, which the NE5532 is and the NE5534 isn't, so choose wisely.
I hope that this spares someone the frustration and futility I have just been through!