login
A370558
Periodic sequence S_1 of length 28: used to compute the number of times a day from 1 to 28 of a month occurs on a day of the week for the Gregorian cycle of 400 years.
3
1, 2, 2, 1, 2, 1, 2, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 3, 1, 1, 2, 2, 1, 2, 1, 2, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 3, 1, 1, 2, 2, 1, 2, 1, 2, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 3, 1, 1, 2, 2, 1, 2, 1, 2, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 3, 1
OFFSET
0,2
COMMENTS
For the Gregorian Calendar (in use since Friday, October 15, 1582) of period legth 400, the number of occurrences of a day d from 1 to 28 of a month on a day D of the week from 0 to 6 (0 for Sunday), can be given for the representative years y fom 0 to 399 (for years congruent to modulo 400, and y >= 1583) by pieces of the present sequence S_1 with period length 28.
These multiplicities are denoted by a length 400 sequence M(d, D) (to be used periodically). It is sufficient to consider the representative days congruent to modulo 7, namely d = 1 representing 8, 15, 22, d = 2 for 9, 16, 27, ...., d = 7 for 14, 21, 28. Then M(d, D), for d = 1, 2, ..., 7 can be reduced further giving only the results for d = 1 for the seven values of D, using the identity M(d, D) = M(1, (D - d' + 1) mod 7), with d' = 0 if d = 7, and d' = d for 1, 2, 3, ..., 6. This builds a 7 X 7 Toeplitz matrix with main diagonal entries 0, upper diagonal values 1 to 6 and lower ones from 6 to 1. E.g., row d = 6: M(6, D) = M(1, j(D)), with j(D) = 2, 3, 4, 5, 6, 0, 1, for D = 0, 1, ..., 6, respectively.
The results for the seven sequences M(1, D) of length 400 are then composed of subsequences of S_1 (to be used periodically). They are given in detail in the linked WL paper.
E.g., the number of times the first day of a month occurs on a Sunday (0), hence the 13th of a month falls on a Friday (5), is given by the length 400 sequence M(1, 0) encoded by (0)101(17)99(17)101(17)99. This translates to: start with S_1(0) = 1 for all together 101 entries (reaching periodically S_1(16) = 1), then take 17 steps to reach the entry S_1(5) = 1, and go on for 99-1 = 98 entries to reach S_1(19) = 2. Then take another 17 steps to reach S_1(8) = 1, and go on for 101-1 = 100 entries to reach S_1(24). Further 17 steps lead to S_1(13) = 2. The next 99-1 entries end up with S_1(27) = 1, closing a cycle. That always jumps with 17 steps appear is a coincidence.
This gives the 4 pieces of M(1, 0) as S_1, S_1, S_1, S_1(0..16); S_1(5..27), S_1, S_1, S_1(0..19); S_1(8..27), S_1, S_1, S_1(0..24); and S_1(13..27), S_1, S_1, S_1. Here S_1(i..j) denotes the subsequence S_1(i), S_1(i+1), ..., S_1(j), i.e., a(i), a(1+1), ..., a(j).
LINKS
Wolfdieter Lang, Four Sequences of Length 28 and the Gregorian Calendar, arXiv:2311.13646v2 [math.HO], 2023.
Index entries for linear recurrences with constant coefficients, signature (1,0,0,-1,1,0,0,-1,1,0,0,-1,1,0,0,-1,1,0,0,-1,1,0,0,-1,1).
FORMULA
{a(n)}_{n>=0} = repeat(1, 2, 2, 1, 2, 1, 2, 2, 1, 3, 1, 1, 3, 2, 1, 3, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 3, 1).
From Chai Wah Wu, Apr 12 2024: (Start)
a(n) = a(n-1) - a(n-4) + a(n-5) - a(n-8) + a(n-9) - a(n-12) + a(n-13) - a(n-16) + a(n-17) - a(n-20) + a(n-21) - a(n-24) + a(n-25) for n > 24.
G.f.: (-x^24 - 2*x^23 + 2*x^22 - x^21 - x^20 - x^19 + 2*x^18 - 2*x^17 - x^16 - x^15 + 2*x^14 - x^13 - 3*x^12 + x^11 + x^10 - 2*x^9 - x^8 + x^7 - x^6 - 2*x^4 + x^3 - x - 1)/(x^25 - x^24 + x^21 - x^20 + x^17 - x^16 + x^13 - x^12 + x^9 - x^8 + x^5 - x^4 + x - 1). (End)
EXAMPLE
M(1, 2) is obtained by the code (4)100(17)101(17)102(17)97, i.e.,
M(1, 2) = S_1(4..27), S_1, S_1, S_1(0..19);
S_1(8..27), S_1, S_1, S_1(0..24); S_1(13..27), S_1, S_1, S_1, S_1(0..2); S_1(19..27), S_1, S_1, S_1, S_1(0..3);
for the four pieces.
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Feb 22 2024
STATUS
approved