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 1-3 of 3 results.

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)))

A231007 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 Gregorian calendar.

Original entry on oeis.org

1, 11, 19, 27, 28, 44, 45, 53, 61, 70, 71, 73, 74, 83, 91, 99, 100, 116, 125, 131, 133, 143, 145, 146, 160, 171, 177, 185, 193, 202, 203, 205, 206, 215, 217, 223, 231, 232, 248, 249, 257, 263, 265, 274, 275, 277, 278, 287, 295, 303, 309, 320, 334, 335, 337, 347, 349, 355, 364
Offset: 1

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.
These are the terms of A231005 not in A231006.

Examples

			1 belongs to this sequence because February 1, 2013 falls on the same day as March 1, 2013, but both February and March do not have the same number of days. In fact, a difference of 1 month can never produce the same calendar for the entire month, with the same number of days.
11 belongs to this sequence because December 1, 2011 falls on the same day as November 1, 2012 but both December and November do not have the same number of days. In fact, a difference of 11 months can never produce the same calendar for the entire month, with the same number of days.

Links

Crossrefs

Cf. A230995-A231014.
Cf. A231012 (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,j=0;for(q=0,4800,if(y[(q%4800)+1]==y[((q+p)%4800)+1],j=1;break));for(q=0,4800,if(y[(q%4800)+1]==y[((q+p)%4800)+1]&&x[(q%4800)+1]==x[((q+p)%4800)+1],j=2;break));if(j==1,print1(p", ")))

A231009 Number of months after which it is not possible to have the same calendar for the entire month with the same number of days, in the Gregorian calendar.

Original entry on oeis.org

1, 2, 4, 5, 7, 10, 11, 12, 13, 16, 19, 21, 24, 25, 27, 28, 30, 33, 36, 39, 42, 44, 45, 47, 48, 51, 53, 56, 59, 61, 62, 65, 70, 71, 73, 74, 79, 82, 83, 85, 88, 91, 93, 96, 97, 99, 100, 102, 105, 108, 111, 116, 119, 120, 125, 128, 131, 133, 134, 137, 139, 142, 143, 145, 146, 148
Offset: 1

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.
This is the complement of A231006.

Links

Crossrefs

Cf. A230995-A231014.
Cf. A231014 (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,j=0;for(q=0,4800,if(y[(q%4800)+1]==y[((q+p)%4800)+1]&&x[(q%4800)+1]==x[((q+p)%4800)+1],j=1;break));if(j==0,print1(p", ")))
Showing 1-3 of 3 results.

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