A231001 Number of years after which an entire year can have the same calendar, in the Julian calendar.

0, 6, 11, 17, 22, 28, 34, 39, 45, 50, 56, 62, 67, 73, 78, 84, 90, 95, 101, 106, 112, 118, 123, 129, 134, 140, 146, 151, 157, 162, 168, 174, 179, 185, 190, 196, 202, 207, 213, 218, 224, 230, 235, 241, 246, 252, 258, 263, 269, 274, 280, 286, 291, 297, 302, 308, 314, 319, 325

Offset: 0

Author

Aswini Vaidyanathan, Nov 02 2013

Keywords

• nonn
• easy

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 a subsequence of A231000.

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Time And Date, Repeating Calendar
• Time And Date, Julian Calendar
• Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,1,-1).

Crossrefs

Cf. A230995-A231014.

Cf. A230996 (Gregorian calendar).

Programs

Mathematica

LinearRecurrence[{1,0,0,0,1,-1},{0,6,11,17,22,28},60] (* Harvey P. Dale, Jun 19 2023 *)

PARI

for(i=0,420,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),print1(i", ");break)))

PARI

concat(0, Vec(x*(2 + 3*x + 2*x^2)*(3 - 2*x + 3*x^2) / ((1 - x)^2*(1 + x + x^2 + x^3 + x^4)) + O(x^40))) \\ Colin Barker, Oct 17 2019

Formula

From Colin Barker, Oct 17 2019: (Start)

G.f.: x*(2 + 3*x + 2*x^2)*(3 - 2*x + 3*x^2) / ((1 - x)^2*(1 + x + x^2 + x^3 + x^4)).

a(n) = a(n-1) + a(n-5) - a(n-6) for n>5.

(End)