Identical Years Table (Reusable Agenda & Calendar)
ccm81 Posted messages 11033 Status Membre -
Is there a table that shows identical years (January 1st starting on the same day of the week, December 31st the same, leap years or not, etc.)?
The goal is to know today which calendars and agendas from the past I can use in the coming years...
I understand that years repeat every 28 years, but there are many exceptions... (9 years, 17 years, etc.) due to leap years, for example, but not only...
My current research is to find out which agenda I can use in 2024. But my question is broader than just one year!
Thank you for your answers.
9 réponses
Hello,
Here's an idea to check, I just made this Excel table: https://www.cjoint.com/doc/23_10/MJgsfYD4O8I_Ann%C3%A9es-identiques.xlsx
In A1 you enter a year, we calculate the day of the week for 01/01 and 31/12 and do the same for the previous years, then we check if they are the same values as the first row.
Hello everyone
For the calendar to be truly identical from one year to the next, Easter must fall on the same day.
Using a table of identical years found online, I calculated the date of Easter for each year in LibreOffice (which has a dedicated function unlike Excel!!!) and highlighted the similar dates.
https://www.cjoint.com/c/MJhoWUqOBvQ
Best regards
Via
Hello,
There is an Excel formula that is easily found to calculate the date of Easter, but that is not the question.
Why Easter and not the variable days that depend on it (Ascension, Pentecost, and the following Monday) or the fixed public holidays whose day of the week they fall on can have an impact depending on the activity?
Hello Brucine
Yes, we can calculate it with Excel using a formula, but there is no DIMANCHEDEPAQUES function like in Libre Office
Why Easter? Just as you said, the other movable feasts depend on it, so the years in which Easter falls on the same date will have Ascension on the same date, etc.
As for the holidays, if January 1st of two years is the same day, then July 14th of those two years will be on the same day, etc.
Thus, in my table, the years that begin the year on a Monday and for which Easter is on Sunday, April 15th are 1979, 1990, 2001, 2063, 2074, and 2085
Hello,
Just ask our friend Google. We have results with more or less explanations:
https://www.slate.fr/story/112075/gardez-calendrier-2015-reutiliser-2026
So in 2024, a leap year, you can use the calendar from 1996. (That’s a bit far!)
What is well conceived is clearly expressed,
And the words to say it come easily.
(Boileau)
Hello everyone
Another simple example, you provide the year in F2
https://www.cjoint.com/c/MJhnJbGFaCB
Best regards
Hello
Taking into account the dates of Easter as noted by via55 (valid from 1900 to 2200)
https://www.cjoint.com/c/MJioRxFCU7B
Regards
Hello everyone! I thank all the participants for their responses, each of which has been studied with seriousness. All of them helped me understand the calculation mechanism.
I have therefore tried to note, year by year, from 2024 to 2059 (I will be 100 years old soon) what were the agendas inherited from my ancestors that I could use in the coming years (this also works for calendars).
Unfortunately, I'm starting with a difficult year: 2024 (next year). Indeed, only my agendas from the years 1940, 1968, and 1996 are identical to 2024... I still need to have one!!
My goal is to classify all my stock, not by the year indicated, but by the year of possible future use. For example, we'll find the first agenda in my pile:
1996 (for 2024), then
2014 or 2003 or 1997 or 1986 or 1975 or 1969... (for 2025), then
2015 or 2009 or 1998 or 1987 or 1981 (for 2026), etc.
I will obviously stack the oldest of my agendas for the same year, if applicable.
This is very complicated, I concede.
On the other hand, I tried to modify the Excel file provided by one of you (jee pee) by attempting to add a macro that would create a new tab in which there would only be the years from column A that have "Yes" in column D. Thus, the similar agendas for a given year would appear. Ideally, I would like to have a table with column A from 2099 to 2024 and columns B, C, D, E, etc. filled with similar years! For example:
2090 | 2079 | 2073 | 2062 | 2051 | 2045 | 2034 | 2023 | 2017 | 2006 | 1995 | 1989 | 1978 | 1967 | 1961 | 1950 | 1939 | 1933
2089 | 2078 | 2067 | 2061 | 2050 | 2039 | 2033 | 2022 | 2011 | 2005 | 1994 | 1983 | 1997 | 1966 | 1955 | 1949 | 1938
...
2024 | 1996 | 1968 | 1940
In my opinion, it should start from the 2nd tab, indicating 2090 in A1. Then the 2nd tab would load 2090 into the 1st tab in A1, and bring 2079 into "tab2"!B1, 2073 into "tab2"!C1, etc. until the year 1940 (I don't have any agendas older than that!)
Is this really feasible?
Even with a macro, I haven't succeeded. Or else I need to go through QuickBasic (1980), but it's been a long time since I programmed that way!!
Thank you for your responses, once again!
Hello,
I misunderstood, I went back to my initial proposal which was in the future by modifying the arithmetic to the past.
The initial table only demonstrates the recurrence depending on whether the year relates to a leap year as 0, -1, -2, or -3.
The only thing that interests you, I don’t know if it suits you, are rows 23 and 24 where I entered 2 consecutive years in B23 and B24.
If I want to project after the first 2 years in B23 and B24, I write in B25
=DATE(YEAR(B24)+1,1,1)
I copy and paste the rest of row 24 to row 25, then from row 26 onwards, I fully copy and paste row 25 and so on.
To refine it, it would be necessary to set a condition so that the date does not fall below 1900 otherwise Excel will crash on us.
https://www.cjoint.com/c/MKbuLsmFKJF
WELL DONE and thank you again VERY MUCH!
I followed your instructions and the result is CORRECT.
I had to modify the syntax of the formula we are repeating on the line because I do not have OFFICE365.
So I put this:
=IFERROR(IF(YEAR(B23)<1940;"-";(IF(DAY(EOMONTH(DATE(YEAR(B23);2;1);0))=29;DATE(YEAR(B23)-28;1;1);IF(DAY(EOMONTH(DATE(YEAR(B23)+1;2;1);0))=29;DATE(YEAR(B23)-6;1;1);DATE(YEAR(B23)-11;1;1)))));"-")
The "iferror" allows nothing to be displayed if the year on the left is already before 1940, which suits me. The dash "-" can also be removed.
Thank you for your input.
I don't have 365 either, only 2019, and mea culpa, I used IF.CONDITIONS because I'm lazy instead of nesting IFs.
Speaking of laziness and not having censored 1900 (or 1940), I didn't do it either if, for a reason that belongs to you, you want to project yourself far into the future: anything that crosses 2100, which is not a leap year, will crash the formula and this condition and the recurrence would need to be reviewed accordingly.
Hello,
The situation can be "mathematized" without much difficulty.
One year out of 4 consecutive years is necessarily a leap year, so from a given year, it is enough to check whether the current year is N to N-3 leap; there will be a shift of one day of the week for each leap year crossed since it is known that for a given year, January 1st and December 31st fall on the same day if not a leap year and increase by one day in the opposite case.
It is noted in line 5 that the iteration depends only on this factor.
The corollary is that if a year is not a leap year, January 1st of the following year rises by one day of the week, and otherwise by two, and for two specific years to have the same days of the week for January 1st and December 31st, both must be leap years or non-leap years.
Therefore, in addition to the famous iteration, it is necessary to check whether the calculated arrival year is of the same type as the starting year and move on to the next if it is not the case.
To complicate matters, the astronomers of Pope Gregory decreed that to catch up with solar time, a year is a leap year if it is divisible by 4 or if it is divisible by both 100 and 400: 2000 is a leap year, but 2100 is not.
The direct formula for calculation is tricky; I only verified whether a given year has 29 days in February, Excel handles this very well, including in 2100, but not my additions of years which logically skip any overlapping date.
The situation can be resolved by considerably complicating the formula, which I did not do on the assumption that we would all be dead by 2100, and therefore it does not work there.
https://www.cjoint.com/c/MJhjBhbcKfF