Issues related to the output

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

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

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 So, nowadays, this IOp is obsolete and -1 should correspond to 0.

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

Yes.

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

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.

1. IOp(3/74)=-5 (see line 8) do not exist in the IOps reference

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

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

1. On line 9 one sees "4//1" What does it mean?

All options have their default value.

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

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