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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 (list; graph; refs; listen; history; text; internal format)
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 2-periodic. - 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(n-1)+28, for all n > 1. - M. F. Hasler, Apr 14 2015

a(2n) = 28n-5 (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*28-5*(-1)^n \\ M. F. Hasler, Apr 14 2015

CROSSREFS

Cf. A230995-A231014.

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

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Last modified August 20 06:32 EDT 2017. Contains 290824 sequences.