
Re: Calendar formula for 2nd Wednesday of each successive month
Posted:
Jan 25, 2013 6:00 PM


On Thu, 24 Jan 2013 20:51:15 0800 (PST), Archimedes Plutonium wrote:
> >The last time I wrote about a calendar curiosity was >when I asked how many calendar years do I need in order to not have to >buy a new calendar. And the answer is 7, if we ignore leap years. The >answer is 7 because I need only 7 calendars that start the january 1st >with one of the seven days of the week. If I have those, I need not >buy any new calendar. > >But now I have a new calendar question, sort of a reversal of the 7 >calendars. I am receiving social security checks every 2nd wednesday >of the month. >So the question is, what math formula can be written that tells me how >many days in each month, starting January of 2013 for the next ten >years, how many days in each month that I have to wait for the check.
The approach of finding and adapting a general formula seems overly complicated for the oneoff task.
It will probably be easier to make a table of dates in a spreadsheet and extract the desired dates.
>For example, January 2013, the first wednesday was 2nd and the second >wednesday was the 9th which means I had to wait 9 days for Jan 2012 to >receive the check. Now Feb 2012, the first wednesday is 6th and the >second wednesday is the 13th so I have to wait 13 days. > >So far I have this: >2013 >Jan wait 9 >Feb wait 13 >. >. >.
Your results are tabular.
A spreadsheet can produce this output.
>So what is the formula that gives me those numbers without consulting >a calendar? Here I would have to include leap years. > >And it is obvious that the numbers have a lower limit of 7 and a upper >limit of 15, depending on what day is the first day of that month. > >What I am interested in is whether there is a internal pattern that >can easily tell me if a month is going to have a early payday or >whether it is near to 15 day wait. > >And I wonder if some years are going to have many 7 day paydays or >many 15 day paydays, given that a >probability of a 7 or 15 day month is about 1 per year since we have >12/7 = 1.7 > >Anyone figure out a formula?
Maybe, but it would be complicated and would probably be evaluated in a spreadsheet.
Why bother with a formula when the spreadsheet can produce the result you want without complexity.
>And I would guess that there is a general formula for what day is the >1st of the month for the next ten years
That would be a list of 120 days.
Such a list is easy to produce in a spreadsheet without needing to find a general forumula.
>has been figured out and that >this formula is part of the solution for the 2nd wednesday of each >month.

