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A231001
Number of years after which an entire year can have the same calendar, in the Julian calendar.
5
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.
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. A230996 (Gregorian calendar).
Sequence in context: A315524 A074993 A315525 * A315526 A315527 A315528
KEYWORD
nonn,easy
AUTHOR
Aswini Vaidyanathan, Nov 02 2013
STATUS
approved