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
KEYWORD
nonn,easy
AUTHOR
Aswini Vaidyanathan, Nov 02 2013
STATUS
approved