Could They Be The Same Day Of The Week?
Python 2, 58 bytes
u=-abs(200-input()%400)-4
print u/100+5>(u-8)*5/4%7>u%4/-3
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A direct formula.
Jelly, 20 18 bytes
99R4ḍj`‘ṡ%4ȷ$S€P7ḍ
Outputs 1 for members, 0 for non-members.
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How it works
99R4ḍj`‘ṡ%4ȷ$S€P7ḍ Main link. Argument: n
99 Set the return value to 99.
R Range; yield [01, .., 99].
4ḍ Test each element in the range for divisibility by 4.
j` Join the resulting array, using itself as separator.
The result is an array of 9801 Booleans indicating whether the
years they represent have leap days.
‘ Increment the results, yielding 1 = 365 (mod 7) for non-leap
years, 2 = 366 (mod 7) for leap years.
%4ȷ$ Compute n % 4000.
ṡ Take all slices of length n % 4000 of the result to the left.
S€ Take the sum of each slice.
P Take the product of the sums.
7ḍ Test for divisibility by 7.
MATL, 17 bytes
`0Gv@+5:YcYO8XOda
The program halts if the input belongs to the sequence, or runs indefinitely (infinite loop) otherwise.
Let n be the input. The code executes a loop that tests years 1 and 1+n; then 2 and 2+n; ... until a matching day of the week is found. If no matching exists the loop runs indefinitely.
The membership function for n is periodic with period 400. Therefore, at most 400 iterations are needed if n belongs to the sequence. This requires less than 20 seconds in Try It Online. As a proof of this upper bound, here's a modified program that limits the number of iterations to 400 (by adding @401<* at the end). Note also that this bound is loose, and a few seconds usually suffice.
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Explanation
` % Do...while
0Gv % Push column vector [0; n], where n is the input number
@+ % Add k, element-wise. Gives [k; k+n]
5: % Push row vector [1, 2, 3, 4, 5]
Yc % Horizontal "string" concatenation: gives the 2×6 matrix
% [k, 1, 2, 3, 4, 5; k+n, 1, 2, 3, 4, 5]. The 6 columns
% represent year, month, day, hour, minute, second
YO % Convert each row to serial date number. Gives a column
% vector of length 2
8XO % Convert each date number to date string with format 8,
% which is weekday in three letters ('Mon', 'Tue', etc).
% This gives a 2×3 char matrix such as ['Wed';'Fri']
d % Difference (of codepoints) along each column. Gives a
% row vector of length 3
a % True if some element is nonzero, or false otherwise
% End (implicit). The loop proceeds with the next iteration
% if the top of the stack is true
Old version, 24 bytes
400:"0G&v@+5:YcYO8XOdavA
Output is 0 if the input belongs to the sequence, or 1 otherwise.
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Explanation
400 % Push row vector [1, 2, ..., 400]
" % For each k in that array
0G&v % Push column vector [0; n], where n is the input number
@+ % Add k, element-wise. Gives [k; k+n]
5: % Push row vector [1, 2, 3, 4, 5]
Yc % Horizontal "string" concatenation: gives the 2×6 matrix
% [k, 1, 2, 3, 4, 5; k+n, 1, 2, 3, 4, 5]. The 6 columns
% represent year, month, day, hour, minute, second
YO % Convert each row to serial date number. Gives a column
% vector of length 2
8XO % Convert each date number to date string with format 8,
% which is weekday in three letters ('Mon', 'Tue', etc).
% This gives a 2×3 char matrix such as ['Wed';'Fri']
d % Difference (of codepoints) along each column. Gives a
% row vector of length 3
a % True if some element is nonzero, or false otherwise
v % Concatenate vertically with previous results
A % True if all results so far are true
% End (implicit). Display (implicit)