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.

A231239 Number of months after which either it is not possible to have a date to fall on the same day of the week, or that it is possible to have a date falling on the same day of the week and the two months have the same number of days, in the Julian calendar.

Original entry on oeis.org

0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, 20, 21, 22, 23, 24, 25, 26, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 42, 43, 44, 46, 47, 48, 50, 51, 52, 53, 54, 55, 56, 57, 59, 60, 62, 63, 64, 65, 67, 68, 69, 70, 72, 75, 76, 77, 78, 79, 80, 81, 82, 84, 85, 86, 88, 89
Offset: 1

Views

Author

Aswini Vaidyanathan, Nov 06 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 A231012.

Links

Crossrefs

Cf. A230995-A231014, A231236-A231239.
Cf. A231238 (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", ")))

A231237 Number of years after which it is either not possible to have a date falling on same day of the week, or the entire year can have the same calendar, in the Julian calendar.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 24, 25, 26, 27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 52, 53, 54, 55, 56, 57, 58, 59, 60, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75
Offset: 1

Views

Author

Aswini Vaidyanathan, Nov 06 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 28.
This is the complement of A231002.

Links

Crossrefs

Cf. A230995-A231014, A231236-A231239.
Cf. A231236 (Gregorian calendar).

Programs

  • PARI
    for(i=0,420,j=0;for(y=0,420,if(((5*(y\4)+(y%4))%7)==((5*((y+i)\4)+((y+i)%4))%7),j=1;break));for(y=0,420,if(((5*(y\4)+(y%4))%7)==((5*((y+i)\4)+((y+i)%4))%7)&&((5*(y\4)+(y%4)-!(y%4))%7)==((5*((y+i)\4)+((y+i)%4)-!((y+i)%4))%7),j=2;break));if(j!=1,print1(i", ")))

Formula

From Chai Wah Wu, Jun 04 2024: (Start)
a(n) = 2*a(n-1) - 2*a(n-2) + 2*a(n-3) - 2*a(n-4) + 2*a(n-5) - 2*a(n-6) + 2*a(n-7) - 2*a(n-8) + 2*a(n-9) - 2*a(n-10) + 2*a(n-11) - 2*a(n-12) + 2*a(n-13) - 2*a(n-14) + 2*a(n-15) - 2*a(n-16) + 2*a(n-17) - 2*a(n-18) + 2*a(n-19) - 2*a(n-20) + 2*a(n-21) - 2*a(n-22) + 2*a(n-23) - 2*a(n-24) + 2*a(n-25) - a(n-26) for n > 26.
G.f.: x^2*(x^4 + 1)*(x^2 - x + 1)*(x^18 + x^17 + x^16 - x^13 + x^10 + x^9 + x^8 - x^5 + x^2 + x + 1)/((x - 1)^2*(x^12 - x^11 + x^10 - x^9 + x^8 - x^7 + x^6 - x^5 + x^4 - x^3 + x^2 - x + 1)*(x^12 + x^11 + x^10 + x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)). (End)

A231238 Number of months after which either it is not possible to have a date to fall on the same day of the week, or that it is possible to have a date falling on the same day of the week and the two months have the same number of days, in the Gregorian calendar.

Original entry on oeis.org

0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, 20, 21, 22, 23, 24, 25, 26, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 46, 47, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 59, 60, 62, 63, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 78, 79, 80, 81, 82, 84, 85, 86, 87
Offset: 1

Views

Author

Aswini Vaidyanathan, Nov 06 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 A231007.

Links

Crossrefs

Cf. A230995-A231014, A231236-A231239.
Cf. A231239 (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", ")))
Showing 1-3 of 3 results.

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