How to determine the number of a given day of the week of a given year using bash date without loops?

Question

bash

Actually the question, I came up with a problem for myself when I was drunk, and for a week now she hasn’t let me go.
How to determine the number of a given day of the week of a year using bash date without loops?
For example, the script takes 2018 6 (the first Sunday of 2018) and outputs 6 (the sixth number).
Takes 2003 3 (the first Thursday of 2003) and outputs 2 (the second number).
Why no cycles?

Because I am a freak

Because that's how I came up with it.
It seemed to sound just like two fingers on the asphalt, and then he sat and turned gray.

Answer 1

Um, the first Sunday seems to be the 7th...
In general, either I misunderstood, or you calculated something incorrectly.
Directly in the console:

year=2018
weekday=6 (суббота)
date -d "$year-01-01-$(date -d "$year--01-01" +%u)days+$weekday days +$((7*$(($year-$(date -d "$year-01-01-$(date -d "$year--01-01" +%u)days +$weekday days" +%Y ))))) days" +%d

At the output "06", i.e. first Saturday 2018)
Count: 1-Monday ...
It also works like this: weekday=6 (Saturday of the first week), weekday=13 (same Saturday, but the second week), 20 (Saturday of the third week), etc.
You owe me my 15 minutes of my life :DDD