Fail with fort.56 with symm=1

Dear MRCC experts,

For a close shelled system it should be output singlet and triplet states, but the code give me states all with spin multiplicity being 2. I found fort.56 always reads nsing=0, ntrip=0 although I have defined symm=1 with nsing and ntrip of some nonzero values. I am using MRCC2015, since I still have some trouble with MRCC2019. Could you help me have a check, thank you very much!

[ymyu@admin1 Be]$ less MINP
iface=dirac
calc=CCSD
mem=7500MB
symm=1
nsing=2
ntrip=9
refdet=vector
1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

and

[ymyu@admin1 Be]$ less fort.56
2 0 0 0 1 0 0 1 0 0 1 0 11 0 0 6 0 0 0.00000 0 7500 0 0.000E+00
ex.lev,nsing,ntrip, rest,CC/CI,dens,conver,symm, diag, CS ,spatial, HF ,ndoub,nacto,nactv, tol ,maxex, sacc, freq, dboc, mem, locno, eps
1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

I attached one test output, thank you very much for your help!

Best regards,
Yanmei

Dear Yanmei,
The file fort.56 is correct. For relativistic calculations ndoub equals the total number of states, and nsing and ntrip should be zero.

Dear Prof. Kallay,

The output giving Spin multiplicity=2 is correct or not? For singlet and triplet state, S equals to 0 and 1, so the spin multiplicity should be 1 and 3? Thank you very much!

Best regards,

Yanmei

That is correct. In the relativistic case, these should be ignored.

Prof. Kallay,

I found this output is different from my old impression using old version of mrcc so felt lost, thank you clarify my confusion.

Best regards,

Yanmei