|
| |
|
|
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.
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
|
|
|
|
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
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)
|
|
|
MATHEMATICA
|
LinearRecurrence[{1, 1, -1}, {5, 23, 33}, 70] (* Harvey P. Dale, May 21 2021 *)
|
|
|
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) Vec(x*(5 + 18*x + 5*x^2) / ((1 - x)^2*(1 + x)) + O(x^50)) \\ Colin Barker, Oct 15 2019
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
