A231014 Number of months after which it is not possible to have the same calendar for the entire month with the same number of days, in the Julian calendar.

1, 2, 4, 5, 7, 10, 11, 12, 13, 16, 19, 21, 24, 25, 27, 28, 30, 33, 36, 39, 41, 42, 44, 45, 47, 48, 49, 50, 51, 53, 56, 58, 59, 61, 62, 65, 66, 67, 70, 71, 73, 74, 76, 79, 82, 83, 84, 85, 87, 88, 90, 91, 93, 96, 97, 99, 100, 102, 104, 105, 107, 108, 111, 113, 116, 119, 120

Offset

1,2

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 336.

This is the complement of A231011.

Links

Table of n, a(n) for n=1..67.

Time And Date, Repeating Months

Time And Date, Julian Calendar

Prog

(PARI) m=[0, 3, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5]; n=[31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]; y=vector(336, i, (m[((i-1)%12)+1]+((5*((i-1)\48)+(((i-1)\12)%4)-!((i-1)%48)-!((i-2)%48))))%7); x=vector(336, i, n[((i-1)%12)+1]+!((i-2)%48)); for(p=0, 336, j=0; for(q=0, 336, if(y[(q%336)+1]==y[((q+p)%336)+1]&&x[(q%336)+1]==x[((q+p)%336)+1], j=1; break)); if(j==0, print1(p", ")))

Crossrefs

Cf. A230995-A231014.

Cf. A231009 (Gregorian calendar).

Sequence in context: A056908 A296344 A060565 * A231009 A133254 A080725

Adjacent sequences: A231011 A231012 A231013 * A231015 A231016 A231017

Keyword

nonn,easy

Author

Aswini Vaidyanathan, Nov 02 2013

Status

approved