[Pw_forum] all electron calculation problem

[Arif Ismail]

Hi,
I'm trying to generate a US-PP for Ce using Vanderbilt's package, and the
first step is to run the AE calculation. This is my input file:

   1    0    0    0    4              ifae,ifpsp,ifprt,ifplw,ilogd (5i5)
   1.80     -2.4       1.6       600  rlogd,emin,emax,nnt (3f10.5,i5)
   1.0d-10   1.0d-09   0.10     600   thresh,tol.damp (2e10.1,f10.5)
 Cerium
  58   0.0       5.0
 200.0        4.0       39.0          rmax,aasf,bbsf (3f10.5)
   15    2                            ncspvs,irel (2i5)
 100  2.      -10660.8
 200  2.      -3450.2
 210  6.      -219.9
 300  2.      -430.0
 310  6.      -350.4
 320  10.     -220.1
 400  2.      -60.0
 410  6.      -30.7
 420  10.     -20.0
 430  1.      -15.0
 500  2.      -10.0
 510  6.      -5.0
 520  1.      -4.0
 600  2.      -3.0
 610  0.      -2.0


I get this error message:

pseudopotential program version  7.3.6   date:  7 - 14 - 2008
========================================================================

 ifae =   1    ifpsp =   0    ifprt =   0    ifplw =   0    ilogd =   4
 rlogd=   1.80000     emin=  -2.40000     emax=   1.60000     nnt=   60
 thresh, tol =   1.000E-10   1.000E-09    damp = 0.100    maxit = 600

          *************************************************************
           Cerium              PBE - GGA           exchange-correlation
          *************************************************************

 z = 58.00    xion = 0.00    exfact =   5.00000    irel=   2
 ncspvs =   15
 rmax = 200.00000    aasf =   4.00000    bbsf =  39.00000

 value of mesh generated in rinit is   523
 irel = 2 so all electron calculations use koelling-harmon equations
 ***error in koesol
 ninf  524 too big for this mesh  523


I tried changing aasf, bbsf, and rmax, and it changes the value of mesh
generated in rinit ... but still I get the same error - ninf is always 1
greater than the mesh. Does anyone know how to solve this problem?
I think it may have to do with the fact that it's a lanthanide atom.
Thanks.