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fwang2011

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¿ÉÄÜÕâ²¢²»½ö½öÊǸöGAMESSµÄÎÊÌ⣬¸üÏñÒ»¸ö»ù±¾µÄÁ¿»¯¼ÆËãÎÊÌâ¡£

ÒÔˮΪÀý£¬if we run the following:
$CONTRL SCFTYP=RHF RUNTYP=OPTIMIZE MAXIT=30 MULT=1 MOLPLT=.TRUE. $END
$SYSTEM TIMLIM=525600 MEMORY=1000000 $END
$BASIS GBASIS=N31 NGAUSS=6 NDFUNC=1 $END
$SCF DIRSCF=.TRUE. $END
$STATPT OPTTOL=0.0001 NSTEP=200 HSSEND=.t. $END
$FORCE METHOD=ANALYTIC VIBANL=.TRUE. $END
$DATA
Test run of H2O
CNV 2

O     8.0     0.00000     0.00000     0.00000
H     1.0    -0.75800     0.00000     0.54500
$END

ÔÚÊä³öÎļþµÄ×îºó£¬we will find the following output in the final MO:
                     11         12         13         14         15
                    1.1785     1.3851     1.4312     2.0206     2.0306
                     A1         B1         A1         A1         A2
    1  O  1  S   -0.051055   0.000000  -0.085666  -0.007482   0.000000
    2  O  1  S   -0.052893   0.000000  -1.443495  -0.078523   0.000000
    3  O  1  X    0.000000  -1.040316   0.000000   0.000000   0.000000
    4  O  1  Y    0.000000   0.000000   0.000000   0.000000   0.000000
    5  O  1  Z    0.731788   0.000000  -0.501204  -0.002830   0.000000
    6  O  1  S    0.411818   0.000000   3.637788   0.234068   0.000000
    7  O  1  X    0.000000   1.540982   0.000000   0.000000   0.000000
    8  O  1  Y    0.000000   0.000000   0.000000   0.000000   0.000000
    9  O  1  Z   -0.315514   0.000000   1.158296   0.130567   0.000000
   10  O  1 XX   -0.249651   0.000000  -0.643896   0.563979   0.000000
   11  O  1 YY    0.080110   0.000000  -0.293083   0.385254   0.000000
   12  O  1 ZZ   -0.063729   0.000000  -0.403630  -1.011299   0.000000
   13  O  1 XY    0.000000   0.000000   0.000000   0.000000   1.000000
   14  O  1 XZ    0.000000   0.026841   0.000000   0.000000   0.000000
   15  O  1 YZ    0.000000   0.000000   0.000000   0.000000   0.000000
   16  H  2  S   -0.694183   0.108435  -0.301211  -0.080583   0.000000
   17  H  2  S    0.368089  -0.934728  -0.812287  -0.037357   0.000000
   18  H  3  S   -0.694183  -0.108435  -0.301211  -0.080583   0.000000
   19  H  3  S    0.368089   0.934728  -0.812287  -0.037357   0.000000

Ò²¾ÍÊÇ˵ÕâÀïÖ»ÓÐ15¸öMO¡£

µ«ÊÇ£¬¼ì²éÒ»ÏÂÎÒÃÇÖªµÀ£º
TOTAL NUMBER OF BASIS SET SHELLS             =    8
NUMBER OF CARTESIAN GAUSSIAN BASIS FUNCTIONS =   19
ÕâÀNUMBER OF CARTESIAN GAUSSIAN BASIS FUNCTIONS¡° µÈÓÚ19£¬Ò²¾ÍÊǽ«ÓÐ19¸ö±¾Õ÷Öµ¡£ÕâÒ²¿ÉÒÔ´ÓÊä³öÎļþÖÐÑéÖ¤¡£

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gmy1990(½ð±Ò+2): 2012-03-04 21:09:38
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2Â¥2012-03-04 10:35:40
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fwang2011

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3Â¥2012-03-04 13:18:50
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fwang2011

½ð³æ (СÓÐÃûÆø)

For more information in this topic, please go to (you have to a member of it):
https://groups.google.com/d/topic/gamess/OBhW8maYlyE/discussion

For those who cannot/don't access the page, I copied the one which I think as answer for your reference.

And the reference is
in statp.scr
C*MODULE STATPT  *DECK SIGEND
       SUBROUTINE SIGEND
line 5359
NMOPRT = MIN(L0,NOCC+10)

in human language min(size of basis, number of occ+10)
min(19,10+5)=15
Robert Balawender
btw Sean

>>In 6-31G(d), 2*H has 2*4 primitives and O has 11 primitives
>> (including
>> the extra d function), giving a total of 19.

19 is a number of the contracted gaussian
and all your answer is wrong

- show quoted text -

--
dr Robert Balawender
Instytut Chemii Fizycznej PAN
ul. Kasprzaka 44/52
01-224 Warszawa
4Â¥2012-05-18 00:57:10
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