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

0,2

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 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).

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)

Mathematica

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

Prog

(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

Crossrefs

Cf. A230995-A231014.

Cf. A230996 (Gregorian calendar).

Sequence in context: A315524 A074993 A315525 * A315526 A315527 A315528

Adjacent sequences: A230998 A230999 A231000 * A231002 A231003 A231004

Keyword

nonn,easy

Author

Aswini Vaidyanathan, Nov 02 2013

Status

approved