cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A231005 Number of months after which a date can fall on the same day of the week, in the Gregorian calendar.

Original entry on oeis.org

0, 1, 3, 6, 8, 9, 11, 14, 15, 17, 18, 19, 20, 22, 23, 26, 27, 28, 29, 31, 32, 34, 35, 37, 38, 40, 41, 43, 44, 45, 46, 49, 50, 52, 53, 54, 55, 57, 58, 60, 61, 63, 64, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 80, 81, 83, 84, 86, 87, 89, 90, 91, 92, 94, 95, 98, 99
Offset: 0

Views

Author

Aswini Vaidyanathan, Nov 02 2013

Keywords

Comments

In the Gregorian calendar, a non-century year is a leap year if and only if it is a multiple of 4 and a century year is a leap year if and only if it is a multiple of 400.
Assuming this fact, this sequence is periodic with a period of 4800.

Examples

			1 belongs to this sequence because February 1, 2013 falls on the same day as March 1, 2013.
3 belongs to this sequence because December 1, 2011 falls on the same day as March 1, 2012.
6 belongs to this sequence because January 1, 2012 falls on the same day as July 1, 2012.
8 belongs to this sequence because March 1, 2013 falls on the same day as November 1, 2013.
9 belongs to this sequence because January 1, 2013 falls on the same day as October 1, 2013.
11 belongs to this sequence because December 1, 2011 falls on the same day as November 1, 2012.

Links

Crossrefs

Cf. A230995-A231014.
Cf. A231010 (Julian calendar).

Programs

  • PARI
    m=[0,3,3,6,1,4,6,2,5,0,3,5];n=[31,28,31,30,31,30,31,31,30,31,30,31];y=vector(4800,i,(m[((i-1)%12)+1]+((5*((i-1)\48)+(((i-1)\12)%4)-((i-1)\1200)+((i-1)\4800)-!((i-1)%48)+!((i-1)%1200)-!((i-1)%4800)-!((i-2)%48)+!((i-2)%1200)-!((i-2)%4800))))%7);x=vector(4800,i,n[((i-1)%12)+1]+!((i-2)%48)-!((i-2)%1200)+!((i-2)%4800));for(p=0,4800,for(q=0,4800,if(y[(q%4800)+1]==y[((q+p)%4800)+1],print1(p", ");break)))

A231011 Number of months after which a date can fall on the same day of the week, and the two months can have the same number of days, in the Julian calendar.

Original entry on oeis.org

0, 3, 6, 8, 9, 14, 15, 17, 18, 20, 22, 23, 26, 29, 31, 32, 34, 35, 37, 38, 40, 43, 46, 52, 54, 55, 57, 60, 63, 64, 68, 69, 72, 75, 77, 78, 80, 81, 86, 89, 92, 94, 95, 98, 101, 103, 106, 109, 110, 112, 114, 115, 117, 118, 123, 124, 126, 127, 129, 132, 135, 140, 141, 147, 149, 150
Offset: 0

Views

Author

Aswini Vaidyanathan, Nov 02 2013

Keywords

Comments

In the Julian calendar, a year is a leap year if and only if it is a multiple of 4 and all century years are leap years.
Assuming this fact, this sequence is periodic with a period of 336.
This is a subsequence of A231010.

Links

Crossrefs

Cf. A230995-A231014.
Cf. A231006 (Gregorian calendar).

Programs

  • PARI
    m=[0,3,3,6,1,4,6,2,5,0,3,5];n=[31,28,31,30,31,30,31,31,30,31,30,31];y=vector(336,i,(m[((i-1)%12)+1]+((5*((i-1)\48)+(((i-1)\12)%4)-!((i-1)%48)-!((i-2)%48))))%7);x=vector(336,i,n[((i-1)%12)+1]+!((i-2)%48));for(p=0,336,for(q=0,336,if(y[(q%336)+1]==y[((q+p)%336)+1]&&x[(q%336)+1]==x[((q+p)%336)+1],print1(p", ");break)))

A231012 Number of months after which a date can fall on the same day of the week, but it is not possible that the two months have the same number of days, in the Julian calendar.

Original entry on oeis.org

1, 11, 19, 27, 28, 41, 45, 49, 58, 61, 66, 71, 73, 74, 83, 87, 91, 100, 104, 113, 121, 130, 131, 133, 138, 143, 146, 159, 160, 176, 177, 190, 193, 198, 203, 205, 206, 215, 223, 232, 236, 245, 249, 253, 262, 263, 265, 270, 275, 278, 287, 291, 295, 308, 309, 317, 325, 335
Offset: 1

Views

Author

Aswini Vaidyanathan, Nov 02 2013

Keywords

Comments

In the Julian calendar, a year is a leap year if and only if it is a multiple of 4 and all century years are leap years.
Assuming this fact, this sequence is periodic with a period of 336.
These are the terms of A231010 not in A231011.

Links

Crossrefs

Cf. A230995-A231014.
Cf. A231007 (Gregorian calendar).

Programs

  • PARI
    m=[0,3,3,6,1,4,6,2,5,0,3,5];n=[31,28,31,30,31,30,31,31,30,31,30,31];y=vector(336,i,(m[((i-1)%12)+1]+((5*((i-1)\48)+(((i-1)\12)%4)-!((i-1)%48)-!((i-2)%48))))%7);x=vector(336,i,n[((i-1)%12)+1]+!((i-2)%48));for(p=0,336,j=0;for(q=0,336,if(y[(q%336)+1]==y[((q+p)%336)+1],j=1;break));for(q=0,336,if(y[(q%336)+1]==y[((q+p)%336)+1]&&x[(q%336)+1]==x[((q+p)%336)+1],j=2;break));if(j==1,print1(p", ")))

A231013 Number of months after which it is not possible to have a date falling on the same day of the week, in the Julian calendar.

Original entry on oeis.org

2, 4, 5, 7, 10, 12, 13, 16, 21, 24, 25, 30, 33, 36, 39, 42, 44, 47, 48, 50, 51, 53, 56, 59, 62, 65, 67, 70, 76, 79, 82, 84, 85, 88, 90, 93, 96, 97, 99, 102, 105, 107, 108, 111, 116, 119, 120, 122, 125, 128, 134, 136, 137, 139, 142, 144, 145, 148, 151, 153, 154, 156, 157, 162
Offset: 1

Views

Author

Aswini Vaidyanathan, Nov 02 2013

Keywords

Comments

In the Julian calendar, a year is a leap year if and only if it is a multiple of 4 and all century years are leap years.
Assuming this fact, this sequence is periodic with a period of 336.
This is the complement of A231010.

Links

Crossrefs

Cf. A230995-A231014.
Cf. A231008 (Gregorian calendar).

Programs

  • PARI
    m=[0,3,3,6,1,4,6,2,5,0,3,5];n=[31,28,31,30,31,30,31,31,30,31,30,31];y=vector(336,i,(m[((i-1)%12)+1]+((5*((i-1)\48)+(((i-1)\12)%4)-!((i-1)%48)-!((i-2)%48))))%7);x=vector(336,i,n[((i-1)%12)+1]+!((i-2)%48));for(p=0,336,j=0;for(q=0,336,if(y[(q%336)+1]==y[((q+p)%336)+1],j=1;break));if(j==0,print1(p", ")))

A233697 Duration in months between successive instances of a given Julian calendar day of the month falling on a specified day of the week.

Original entry on oeis.org

3, 3, 14, 3, 6, 8, 1, 8, 6, 8, 9, 6, 3, 14, 3, 3, 8, 9, 9, 8, 9, 11, 3, 6, 8, 1, 8, 9, 14, 6, 3, 14, 3, 6, 8, 6, 9, 8, 9, 6, 3, 11, 8, 1, 8, 9, 9, 8
Offset: 1

Views

Author

Adam Helman, Dec 14 2013

Keywords

Comments

A 28-year cycle exists as derived by multiplication of 4 years / leap cycle by 7 days of the week. Within such a cycle there are exactly 48 days when a given day of the month (selected from 1 to 28) falls on a specific day of the week.
Fewer days result for the choices of 29, 30 and 31, each with their own unique sequence that repeats after 28 years. These results may be published as their own related yet nonetheless unique sequences.
The average duration between such events is simply 7 months as trivially derived. However, the *actual* durations vary from 1 to 14 months with the following frequencies:
separation (months) number of instances
1 3
3 10
6 8
8 12 (modal value)
9 9
11 2
14 4
The entire sequence of 48 time separations in months between successive instances of a given day of the month falling on a specified day of the week is provided above.
(Account is not taken of the extra non-leap years introduced with the Gregorian calendar; i.e., those divisible by 100 yet NOT by 400, e.g., A.D. 2100, 2200 and 2300. Thus, the method in the example does not work without modification for time durations including such years or years such as A.D. 2201, 2202 or 2203; in these cases, the previous leap year is 4, 5, 6 or 7 years earlier.)

Examples

			**** THIS FULL EXPLANATION IS REQUIRED TO PROVIDE AN EXAMPLE ****
Designate "year 0" as any leap year, with the succeeding years as years 1, 2 and 3. The first term, 3, is the time in months between a date in January 0 (such as January 17, 2012 - a leap year) and the next time ANY "17th" falls on the same day of the week - April 17, 2012.
The following tabulation identifies the remaining 47 month/year combinations within the sequence, with "year" being 0 through 3 as defined above.
Jan0- 3 -Apr0-  3 -Jul0- 14 -Sep1-  3 -Dec1-  6 -Jun2-  8 -Feb3
Feb3- 1 -Mar3-  8 -Nov3-  6 -May0-  8 -Jan1-  9 -Oct1-  6 -Apr2
Apr2- 3 -Jul2- 14 -Sep3-  3 -Dec3-  3 -Mar0-  8 -Nov0-  9 -Aug1
Aug1- 9 -May2-  8 -Jan3-  9 -Oct3- 11 -Sep0-  3 -Dec0-  6 -Jun1
Jun1- 8 -Feb2-  1 -Mar2-  8 -Nov2-  9 -Aug3- 14 -Oct0-  6 -Apr1
Apr1- 3 -Jul1- 14 -Sep2-  3 -Dec2-  6 -Jun3-  8 -Feb0-  6 -Aug0
Aug0- 9 -May1-  8 -Jan2-  9 -Oct2-  6 -Apr3-  3 -Jul3- 11 -Jun0
Jun0- 8 -Feb1-  1 -Mar1-  8 -Nov1-  9 -Aug2-  9 -May3-  8 -Jan0
**** EXAMPLE ****
It is December 13, 2013 - and a Friday. When are the next 5 "Friday the 13ths"?
SOLUTION. 2013 is a "year 1". Locate December for year 1 in the table in the fifth column of months, first entry. By inspection it is 6 months to the next Friday the 13th in June 2014. Reading the next four entries, the remaining four events are in February 2015 (as "Feb3"), March 2015 ("Mar3"), November 2015 ("Nov3") and May 2016 ("May0").

Links

Crossrefs

Cf. A231010, A231013.

Programs

  • PARI
    {a(n, list = 0) = if(n<1 || n>48, return);
                                   \\ calculate (\list) up to n-th term
my(daysmod7 = [3, 0, 3, 2, 3, 2, 3, 3, 2, 3, 2, 3],
                                   \\ days beyond 28 in months
s = 0, c = 1, yr = 0, mo = 1, months = 0);
                                   \\ yr = 0 means (Julian) leap year
while(1,                           \\ find next cumulative whole week(s)
  months++; s = (s + daysmod7[mo] + (mo==2 && yr==0))%7;
                                   \\ count elapsed months
  if(s==0, if(list, print1(months, ", "));
                                   \\ print if optional list <> 0
           if(c==n, return(months), c++; months = 0));
                                   \\ return n-th or continue
  if(mo<12, mo++, mo = 1; yr = (yr + 1)%4))}
                                   \\ new month or new month/year
a(48, 1) \\ print all 48 terms (and return 48th) - Rick L. Shepherd, Jul 06 2017

Formula

Brute force calculation is more efficient than employing a formula, one that would be most involved and hence not meaningful in its individual terms. Calendar consultation provides a simple alternative.

Extensions

Lightly edited and terms/examples checked by Rick L. Shepherd, Jun 29 2017
Showing 1-5 of 5 results.

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