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For example, for $$w = 2^ − 1$$ and for $$f$$ and $$d$$ near $$f ≈ 25920$$ and $$d ≈ 765433$$ there is no solution with $$X ≥ d$$ for only 1 out of about 9100 combinations of $$f$$ and $$d$$, and no solution with $$X ≥ w$$ for only 1 out of about 54 combinations.

However, when $$f$$ and $$d$$ are considerably less than $$w$$, then there is a good chance of a solution.

For $$z = 0$$ the equation becomes \begin x = \ceilratio − 1 \end and that is certainly correct for $$y$$ for which $$z = 0$$ (i.e., for the first day of each month). \begin \begin x & = \ceilratio − 1 \ & = \ceilratio − 1 \ & = \ceilratio − 1 \ & = x − 1 \ceilratio \end \ \begin 1 & = \ceilratio \ 0 & \lt \frac ≤ 1 \ 0 & \lt −⌊px r⌉_1 z 1 ≤ p \ ⌊px r⌉_1 − 1 & \lt z ≤ p − 1 ⌊px r⌉_1 \ 0 & ≤ z ≤ ⌊p⌋ − 1 ⌊p⌉_1 ⌊px r⌉_1 \end \end So, $$z$$ may be as large as $$⌊p⌋ − 1$$ (i.e., at least $$⌊p⌋$$ days in the month) for any month, and may be as large as $$⌊p⌋$$ (i.e., $$⌊p⌋ 1$$ days in the month) if $$⌊p⌉_1 ⌊px r⌉_1 ≥ 1$$, which is exactly the condition we found earlier for month $$x$$ to be a long month, so it all fits together.

Then \begin \begin c(x) & = c(x_0 a) − c(a) \ & = ⌊p(x_0 a)⌋ − ⌊pa⌋ \ & = ⌊px_0 pa − ⌊pa⌋⌋ \ & = ⌊px_0 ⌊pa⌉_1⌋ \end \end so \begin r = ⌊pa⌉_1 \end The subtracting of $$c(a)$$ ensures that $$c(0) = 0$$. \begin \begin y & = ⌊px r⌋ z \ px r z − 1 & \lt y ≤ px r z \ x & \lt \frac ≤ x \frac ≤ x 1 \end \end $$x$$ must be a whole number, and there is only one whole number that can satisfy the last line of inequalities, and that is \begin x = \ceilratio − 1 \end To use the above equation we must know the value of $$z$$, but in general we don't know that value yet.

For a given $$w$$ there are many combinations of $$d$$ and $$f$$ for which there are no solutions with $$X ≥ w$$ or $$X ≥ d$$.

If such a solution does not exist, then we seek a solution for $$X ≥ d$$ (for which we do need to use equations \eqref) and then we must try up to 86 million values of $$R$$, which is still much less than the up to 1100 million values of $$P$$.

The formulas for $$\max(R_d)$$, $$N_w$$, and $$N_d$$ are approximations; the other formulas are exact for $$10 ≤ w ≤ 1024$$.

I find \begin \max(f_w) & ≈ \floorratio \ \max(f_d) & ≈ \floorratio \ \max(d_w) & ≈ w − 5 (w \bmod 2) \ \max(d_d) & ≈ w − 1 \ \max(c_w) & ≈ \floorratio \ \max(c_d) & ≈ \floorratio \ \max(P_w) & ≈ \floorratio − 2 (w \bmod 2) \ \max(P_d) & ≈ \floorratio − 1 \ \max(Q) & ≈ \floorratio \ \max(R_w) & ≈ \floorratio \ \max(R_d) & ≈ \frac \ \max(X) & ≈ \floorratio \ N_w & ≈ \fracw^ \ N_d & ≈ \fracw^ \end where the variables with subscript $$w$$ apply only to $$X \gt w$$, those with subscript $$d$$ apply only to $$X \gt d$$, and those without a subscript apply to both.

The Extended Algorithm of Euclid (or a search) yields that $$−15188f 499d = 3$$, and there is no sum of a multiple of $$f$$ and a multiple of $$d$$ that is closer to 0.

Suppose that that pattern would be good if only it were shifted by $$a$$ months so that month $$x_0$$ in the calendar with $$r = 0$$ corresponded with month $$x = x_0 a$$ in the new calendar.

With $$r = 0$$ we get a certain pattern of long and short months.

On average there is a long month after every \begin Q = \frac \label \end months, but that isn't always a whole number, so in practice the number of months between two successive long months is equal to $$⌊Q⌋$$ or $$⌈Q⌉$$.

Month $$x$$ is a long month (with $$⌊p⌋ 1$$ days) if $$⌊px r⌉_1 ⌊p⌉_1 ≥ 1$$.

Now we calculate $$y$$, $$e$$ for $$x = 1710321$$.