A231238 Number of months after which either it is not possible to have a date to fall on the same day of the week, or that it is possible to have a date falling on the same day of the week and the two months have the same number of days, in the Gregorian calendar.

0, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, 20, 21, 22, 23, 24, 25, 26, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 46, 47, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 59, 60, 62, 63, 64, 65, 66, 67, 68, 69, 72, 75, 76, 77, 78, 79, 80, 81, 82, 84, 85, 86, 87

Offset

1,2

Comments

In the Gregorian calendar, a non-century year is a leap year if and only if it is a multiple of 4 and a century year is a leap year if and only if it is a multiple of 400.

Assuming this fact, this sequence is periodic with a period of 4800.

This is the complement of A231007.

Links

Table of n, a(n) for n=1..74.

Time And Date, Repeating Months

Time And Date, Gregorian 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(4800, i, (m[((i-1)%12)+1]+((5*((i-1)\48)+(((i-1)\12)%4)-((i-1)\1200)+((i-1)\4800)-!((i-1)%48)+!((i-1)%1200)-!((i-1)%4800)-!((i-2)%48)+!((i-2)%1200)-!((i-2)%4800))))%7); x=vector(4800, i, n[((i-1)%12)+1]+!((i-2)%48)-!((i-2)%1200)+!((i-2)%4800)); for(p=0, 4800, j=0; for(q=0, 4800, if(y[(q%4800)+1]==y[((q+p)%4800)+1], j=1; break)); for(q=0, 4800, if(y[(q%4800)+1]==y[((q+p)%4800)+1]&&x[(q%4800)+1]==x[((q+p)%4800)+1], j=2; break)); if(j!=1, print1(p", ")))

Crossrefs

Cf. A230995-A231014, A231236-A231239.

Cf. A231239 (Julian calendar).

Sequence in context: A369092 A247801 A004775 * A231239 A335523 A374519

Adjacent sequences: A231235 A231236 A231237 * A231239 A231240 A231241

Keyword

nonn,easy

Author

Aswini Vaidyanathan, Nov 06 2013

Status

approved