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