Issues related to the output

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Structure of program execution

At the beginning of the output file there is a section describing the structure of the program execution. The numbers refer to which IOps that will be executed an in which order. A part from that, my knowledge on the topic is quite limited. The example below is from an ONIOM calculation and further down I have put some questions I would like you to answer.

1/6=70,8=5,10=4,14=-1,18=1000120,38=1,52=2,56=1,57=2,64=2/1,3;
 2/9=110,15=1,17=6,18=5,40=1/2;
 1/6=70,8=5,10=4,14=-1,18=1000120,38=1,52=2,53=3172,64=2/20;
 3/5=26,6=2,11=9,16=1,25=1,30=1/1;
 4/20=10,22=2,24=3,28=10/2;
 7/7=1,25=1,30=1,33=-1,44=-1/16;
 1/6=70,8=5,10=4,14=-1,18=1000120,52=2,53=2032/20;
 3/5=7,11=2,16=1,17=8,25=1,30=1,74=-5/1,2,3;
 4//1;
 5/5=2,6=10,38=5,94=2/2;
 8/6=4,10=90,11=11/1;
 11/6=1,8=1,9=11,15=111,16=1,31=1/1,2,10;
 10/6=1,7=6,31=1/2;
 6/7=2,8=2,9=2,10=2,28=1/1;
 7/7=1,10=1,18=20,25=1,30=1,33=-1/1,2,3,16;
 1/6=70,8=5,10=4,14=-1,18=1000120,52=2,53=1022,64=2/20;
 3/5=26,6=2,11=9,16=1,25=1,30=1/1;
 4/20=10,22=2,24=3,28=10/2;
 7/7=1,25=1,30=1,33=-1,44=-1/16;
 1/6=70,8=5,10=4,14=-1,18=1000120,52=2,53=3016,64=2/20;
 7/9=1,25=1,44=-1/16;
 1/6=70,8=5,10=4,14=-1,18=1000020,52=2,64=2/3(1);
 99//99;
 2/9=110,15=1/2;
 1/6=70,8=5,14=-1,18=1000020,52=2,53=3173,64=2/20;
 3/5=26,6=2,11=9,16=1,25=1,30=1/1;
 4/16=2,20=10,22=1,24=3,28=10/2;
 7/7=1,30=1,33=-1,44=-1/16;
 1/6=70,8=5,14=-1,18=1000020,52=2,53=2033/20;
 3/5=7,6=1,11=2,16=1,17=8,25=1,30=1,74=-5,82=7/1,2,3;
 4/5=5,16=3/1;
 5/5=2,6=10,38=5,94=2/2;
 7/7=1,30=1,33=-1/1,2,3,16;
 1/6=70,8=5,14=-1,18=1000020,52=2,53=1023,64=2/20;
 3/5=26,6=2,11=9,16=1,25=1,30=1/1;
 4/16=2,20=10,22=1,24=3,28=10/2;
 7/7=1,30=1,33=-1,44=-1/16;
 1/6=70,8=5,14=-1,18=1000020,52=2,53=3015,64=2/20;
 7/9=1,44=-1/16;
 1/6=70,8=5,14=-1,18=1000020,52=2,64=2/3(-16);
 2/9=110,15=1/2;
 99//99;

Questions regarding the above example

Question: # IOp(1/14)=-1 -1 is not mentioned as an option in the IOps reference

Answer: I'm not sure, but Berny algorithms by default follow a linear minimization. So, nowadays, this IOp is obsolete and -1 should correspond to 0.

Question: # IOp(1/18)=1000120 (Does this mean the sum of three options?)

Answer: Yes.

Question: # At the end of line 1 there is a "/1,3;" What does this mean?

Answer: After the second slash are the list (in order) of the links to be executed. For instance, will be executed the links 101 and 103.

Question: # IOp(3/74)=-5 (see line 8) do not exist in the IOps reference

Answer: http://www.gaussian.com/g_tech/overlay_3.htm#3_74

"-5 Becke3 using VWN/LYP for correlation."

Question: # On line 9 one sees "4//1" What does it mean?

Answer: All options have their default value.

Question: # On some lines there is a number in parenthesis at the end. Could someone explain what this mean?

Answer: This is a "jump number". For example (1) means "jump to the next route line" or (-2) means "jump 2 lines before" (to repeat or skip lines).