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.

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

Offset: 1

Author

Aswini Vaidyanathan, Nov 02 2013

Keywords

• nonn
• easy

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

• Colin Barker, Table of n, a(n) for n = 1..1000
• Time And Date, Repeating Calendar
• Time And Date, Julian Calendar
• Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Crossrefs

Cf. A230995-A231014.

Cf. A230997 (Gregorian calendar).

Programs

Mathematica

LinearRecurrence[{1,1,-1},{5,23,33},70] (* Harvey P. Dale, May 21 2021 *)

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

PARI

Vec(x*(5 + 18*x + 5*x^2) / ((1 - x)^2*(1 + x)) + O(x^50)) \\ Colin Barker, Oct 15 2019

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

From Colin Barker, Oct 15 2019: (Start)

G.f.: x*(5 + 18*x + 5*x^2) / ((1 - x)^2*(1 + x)).

a(n) = a(n-1) + a(n-2) - a(n-3) for n>3.

a(n) = -7 + 2*(-1)^n + 14*n.

(End)