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See the end</div>
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So, you are saying that from this file</p>
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Title <br>
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P LATTICE,NONEQUIV.ATOMS: 2 <br>
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MODE OF CALC=RELA unit=ang <br>
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10.263863 10.263863 10.263863 55.256000 55.256000 55.256000<br>
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ATOM -1: X=0.35500000 Y=0.35500000 Z=0.35500000<br>
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ATOM -1:X= 0.64500000 Y=0.64500000 Z=0.64500000<br>
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ATOM -1:X= 0.85500000 Y=0.85500000 Z=0.85500000<br>
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ATOM -1:X= 0.14500000 Y=0.14500000 Z=0.14500000<br>
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Fe NPT= 781 R0=0.00005000 RMT= 2.0000 Z: 26.000<br>
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ATOM 2: X=0.94499999 Y=0.55500000 Z=0.25000000<br>
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ATOM 2:X= 0.05500001 Y=0.44500000 Z=0.75000000<br>
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ATOM 2:X= 0.55500000 Y=0.25000000 Z=0.94499999<br>
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ATOM 2:X= 0.44500000 Y=0.75000000 Z=0.05500001<br>
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ATOM 2:X= 0.25000000 Y=0.94499999 Z=0.55500000<br>
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ATOM 2:X= 0.75000000 Y=0.05500001 Z=0.44500000<br>
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O NPT= 781 R0=0.00010000 RMT= 2.0000 Z: 8.000</div>
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After splitting and symmetrizing, I will arrive at this</p>
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<span style="font-family:Tahoma; font-size:13px">R LATTICE,NONEQUIV.ATOMS: 3 161_R3c <br clear="none">
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<div style="font-family:Tahoma; font-size:13px"> 9.519303 9.519303 26.005203 90.000000 90.000000120.000000 <br clear="none">
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<div style="font-family:Tahoma; font-size:13px">ATOM -1: X=0.14500000 Y=0.14500000 Z=0.14500000<br clear="none">
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<div style="font-family:Tahoma; font-size:13px">Fe1 NPT= 781 R0=0.00005000 RMT= 1.97 Z: 26.000 <br clear="none">
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<div style="font-family:Tahoma; font-size:13px">ATOM -2: X=0.05500001 Y=0.75000000 Z=0.44500000<br clear="none">
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<div style="font-family:Tahoma; font-size:13px">O NPT= 781 R0=0.00010000 RMT= 1.69 Z: 8.000 <br clear="none">
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<div style="font-family:Tahoma; font-size:13px">ATOM -3: X=0.35500000 Y=0.35500000 Z=0.35500000<br clear="none">
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<div style="font-family:Tahoma; font-size:13px">Fe2 NPT= 781 R0=0.00005000 RMT= 1.97 Z: 26.000</div>
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Now, oxygen had multiplicity 6 and now it has 1. Is this correct? Also</p>
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The original cell has 10 atoms and this one has much less and it is quite different from the original.</p>
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<div dir="ltr">The crystal is SG 161, I only put the fist position.</div>
<div dir="ltr">There are two Fe1 (up), two Fe2 (dn) and 6 O <br>
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<div dir="ltr">I started with the 'P' lattice (above) and from the 4 Fe I took the 1st and 3th as Fe2 and 2nd and 4th as Fe1</div>
<div dir="ltr">You can play other combinations, but this seems the best; each O has two Fe1and two Fe2 forming a cross.</div>
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<div dir="ltr">Pablo<br>
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