A231004 Number of years after which it is not possible to have the same calendar for the entire year, in the Julian calendar.

1, 2, 3, 4, 5, 7, 8, 9, 10, 12, 13, 14, 15, 16, 18, 19, 20, 21, 23, 24, 25, 26, 27, 29, 30, 31, 32, 33, 35, 36, 37, 38, 40, 41, 42, 43, 44, 46, 47, 48, 49, 51, 52, 53, 54, 55, 57, 58, 59, 60, 61, 63, 64, 65, 66, 68, 69, 70, 71, 72, 74, 75, 76, 77, 79, 80, 81, 82, 83, 85, 86, 87

Offset

1,2

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 A231001.

Links

Table of n, a(n) for n=1..72.

Time And Date, Repeating Calendar

Time And Date, Julian Calendar

Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,-1).

Formula

From Chai Wah Wu, Jun 04 2024: (Start)

a(n) = a(n-1) + a(n-23) - a(n-24) for n > 24.

G.f.: x*(x^23 + x^22 + x^21 + x^20 + x^19 + 2*x^18 + x^17 + x^16 + x^15 + 2*x^14 + x^13 + x^12 + x^11 + x^10 + 2*x^9 + x^8 + x^7 + x^6 + 2*x^5 + x^4 + x^3 + x^2 + x + 1)/(x^24 - x^23 - x + 1). (End)

Prog

(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)&&((5*(y\4)+(y%4)-!(y%4))%7)==((5*((y+i)\4)+((y+i)%4)-!((y+i)%4))%7), j=1)); if(j==0, print1(i", ")))

Crossrefs

Cf. A230995-A231014.

Cf. A230999 (Gregorian calendar).

Sequence in context: A039224 A161508 A039264 * A039161 A032797 A001967

Adjacent sequences: A231001 A231002 A231003 * A231005 A231006 A231007

Keyword

nonn,easy

Author

Aswini Vaidyanathan, Nov 02 2013

Status

approved