Which one give me the maximum resolution from 1ksps to 10ksps = conversion time of 1mS to 100uS ? LDC1000 or LDC1614

I checked the documentation of the LDC1000 and LDC1614. 

But, I can figure out which one I should choose. 

I want to check inductance at sampling rate around 1ksps to 10ksps. 

(1) LDC1000

www.ti.com/.../ldc1000.pdf

page 24 say

SR = 3*fsensor/(RESPONSE_TIME)

If fsensor = 2MHz

SR = 3*2*10^6/(RESPONSE_TIME)

If RESPONSE_TIME = 192
SR = 3*2*10^6/(192) = 31.25 ksps
tcx = 32 uS

If RESPONSE_TIME = 384
SR = 3*2*10^6/(384) = 15.625 ksps
tcx = 64 uS

If RESPONSE_TIME = 768
SR = 3*2*10^6/(384) = 15.625 ksps
tcx = 128 uS

If RESPONSE_TIME = 1536
SR = 3*2*10^6/(384) = 15.625 ksps
tcx = 256 uS

If RESPONSE_TIME = 3072
SR = 3*2*10^6/(384) = 15.625 ksps
tcx = 512 uS

If RESPONSE_TIME = 6144
SR = 3*2*10^6/(6144) = 976 sps
tcx = 1.024 mS

How many ENOB I can get from each configuration?

(2) LDC1614

www.ti.com/.../ldc1612.pdf
page 13 say

The conversion time represents the number of reference clock cycles used to measure the sensor frequency. It is
set by the CHx_RCOUNT register for the channel. The conversion time for any channel x is:
tCx = (CHx_RCOUNT * 16 + 4) /fREFx (6)
The reference count value must be chosen to support the required number of effective bits (ENOB). For example, if an ENOB of 13 bits is required, then a minimum conversion time of 2
13 = 8192 clock cycles is required. 8192 clock cycles correspond to a CHx_RCOUNT value of 0x0200.

So,

If ENOB = 13 bit
2^13 = 8192 = 0x2000h
CHx_RCOUNT = (2^ENOB)/16 = 2^9 = 512 = 0x0200
tcx = ((CHx_RCOUNT*16+4)/(40*10^6))
tcx = ((((2^ENOB)/16)*16+4)/(40*10^6))
tcx = (((2^ENOB)+4)/(40*10^6))
tcx = (((2^13)+4)/(40*10^6))*10^6 = 204.9 uS (sample value)

If ENOB = 11 bit
2^11 = 2048 = 0x0800
CHx_RCOUNT = (2^ENOB)/16 = 2^9 = 128 = 0x0080 (LDC1312/LDC1614 power on reset value)
tcx = (((2^ENOB)+4)/(40*10^6))
tcx = (((2^11)+4)/(40*10^6))*10^6 = 51.3 uS

If ENOB = 16 bit
tcx = (((2^16)+4)/(40*10^6))*10^3 = 1.6385 mS

By the way, CHx_RCOUNT is 16 bit, so the maximum value is
tcx = ((CHx_RCOUNT*16+4)/(40*10^6)
tcx = ((2^16)*16+4)/(40*10^6)*10^3 = 26.2145 mS
tcx = ((2^20)+4)/(40*10^6)*10^3 = 26.2145 mS ENOB = 20 bit max ?

I'm confused!