

A231002


Number of years after which it is possible to have a date falling on same day of the week, but the entire year does not have the same calendar, in the Julian calendar.


3



5, 23, 33, 51, 61, 79, 89, 107, 117, 135, 145, 163, 173, 191, 201, 219, 229, 247, 257, 275, 285, 303, 313, 331, 341, 359, 369, 387, 397, 415, 425, 443, 453, 471, 481, 499, 509, 527, 537, 555, 565, 583, 593, 611, 621, 639, 649, 667, 677, 695, 705, 723, 733, 751, 761, 779, 789
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OFFSET

1,1


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.
These are the terms of A231000 not in A231001.
The statement about the period is misleading: this is the sequence of (positive) numbers congruent to 5 or 5 (mod 28). It is strictly increasing, not periodic; the sequence a(n)  28*floor(n/2) is 2periodic.  M. F. Hasler, Apr 14 2015


LINKS

Table of n, a(n) for n=1..57.
Time And Date, Repeating Calendar
Time And Date, Julian Calendar


FORMULA

a(n+1) = a(n1)+28, for all n > 1.  M. F. Hasler, Apr 14 2015
a(2n) = 28n5 (n>0), a(2n+1) = 28n+5 (n>=0), a(n) = 28*floor(n/2)5*(1)^n.  M. F. Hasler, Apr 14 2015


PROG

(PARI) for(i=0, 420, j=0; for(y=0, 420, if(((5*(y\4)+(y%4))%7)==((5*((y+i)\4)+((y+i)%4))%7), j=1; break)); 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), j=2; break)); if(j==1, print1(i", ")))
(PARI) A231002(n) = n\2*285*(1)^n \\ M. F. Hasler, Apr 14 2015


CROSSREFS

Cf. A230995A231014.
Cf. A230997 (Gregorian calendar).
Sequence in context: A071199 A082283 A238195 * A050906 A195974 A098421
Adjacent sequences: A230999 A231000 A231001 * A231003 A231004 A231005


KEYWORD

nonn,easy


AUTHOR

Aswini Vaidyanathan, Nov 02 2013


STATUS

approved



