Excel Formula

Dele65 

Just in case there is an award for the longest formula, one could have

= LET(
    startDate,   DATE(year, 1, 1),
    endDate,     EOMONTH(startDate, 11),
    listDates,   SEQUENCE(1 + endDate - startDate, 1, startDate),
    offset,      WEEKDAY(startDate, 3),
    offsetDates, IF(offset, VSTACK(EXPAND("", offset, , ""), listDates), listDates),
    calendar,    WRAPROWS(offsetDates, 7, ""),
    calendar
  )

The start date is given.  The end date is calculated to allow for leap years.  This allows every date of the year to be generated as a single list.  Because I want every week to start with a Monday, I used the weekday to offset the dates with a number of blank cells [0 was a bit of a problem].  WRAPROWS changed the list of days to a calendar display of weeks.  

My objective is to generate long formulas (no 'rule of thumb' for me!) but with each line being as simple as I can make it.

Dele65 

I believe you'd have to add a 53rd row to show 12/31/2023.  If you're building this formula to accommodate future years, 2024 has a leap year so the 53rd row would include (2) 2023 dates.

Try this:

=LET(y,2023,start,DATE(y,1,1),SEQUENCE(53,7,start))

Dele65 

Without knowing exactly what you mean by "produce a 2023 calendar" I suggest that you review this link, from which I've taken the following screen shot

It appears you  are wanting

• 52 rows, one for each week;
• 7 columns, one for each day of the week
• but you've not entered a meaningful start date, nor a step

The correct start would be the value of Jan 1 2023, which happens to be 44927. And then the step is 1

So the correct formula would read =SEQUENCE(52,7,44927,1)

An alternative could be=SEQUENCE(52,7,DATEVALUE("1/1/2023"),1)

At the end of the process, when all the fields of 52 rows and 7 columns are filled, make sure they're all formatted as dates, in whichever of the several possibilities you desire.