A231000 Number of years after which a date can fall on the same day of the week, in the Julian calendar.

0, 5, 6, 11, 17, 22, 23, 28, 33, 34, 39, 45, 50, 51, 56, 61, 62, 67, 73, 78, 79, 84, 89, 90, 95, 101, 106, 107, 112, 117, 118, 123, 129, 134, 135, 140, 145, 146, 151, 157, 162, 163, 168, 173, 174, 179, 185, 190, 191, 196, 201, 202, 207, 213, 218, 219, 224, 229, 230, 235

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.

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 (1,0,0,0,0,0,1,-1).

Crossrefs

Cf. A230995-A231014.

Cf. A230995 (Gregorian calendar).

Programs

Magma

R:=PowerSeriesRing(Integers(), 60); [0] cat Coefficients(R!( x*(1-x+x^2)*(5+6*x+6*x^2+6*x^3+5*x^4)/((1-x)^2*(1+x+x^2+x^3+x^4+x^5+x^6)) )); // Marius A. Burtea, Oct 17 2019

PARI

for(i=0,420,for(y=0,420,if(((5*(y\4)+(y%4))%7)==((5*((y+i)\4)+((y+i)%4))%7),print1(i", ");break)))

PARI

concat(0, Vec(x*(1 - x + x^2)*(5 + 6*x + 6*x^2 + 6*x^3 + 5*x^4) / ((1 - x)^2*(1 + x + x^2 + x^3 + x^4 + x^5 + x^6)) + O(x^40))) \\ Colin Barker, Oct 17 2019

Formula

From Colin Barker, Oct 17 2019: (Start)

G.f.: x*(1 - x + x^2)*(5 + 6*x + 6*x^2 + 6*x^3 + 5*x^4) / ((1 - x)^2*(1 + x + x^2 + x^3 + x^4 + x^5 + x^6)).

a(n) = a(n-1) + a(n-7) - a(n-8) for n>7.

(End)