A 365-day year contains 52.14286 weeks, having one small dry lump in the batter for one day excess, whereas a 366-day year has 52.28571 weeks and two dry lumps. Therefore even the 52-week years contain errors that just didn’t rise to the level of adding an additional week to the annum.

(edit: Also Each year begins on a different day that splits the werk partially to the prior year with remainder to current year. I presume there is an accepted procedure for declaring a 53-week year or it simply might be a 53rd batch if reports is recorded by year-end. In any case, it seems to me that all years have some days of error due to the residual days.)

One thought is whether employment of a resampling algorithm may provide benefit. A first approach is to resample each 53-week year to a 52-week year (or all 52-week yeara to a 53).

A second approach is to resample each year to days, then choose a convenient integer for a unit of time with which to do comparisons (more on that further down).

There are theoreticallly four resamplings to consider - 52 week data for a 365-day year, 52-week data for a 366-day year, and two cases for 53-week years. Resampling ratios: 52/365, 52/366, 53/365, 53/366.

N.B. the 366-day years could be resampled to a 365 day year as a way of neglecting the extra day in each leap year.

The result would be a dataset of resampled days for each year. Next, regrouping is in order to make comparisons tastier.

Choosing 7 re-sampled days for a comparison ’week’ would seem to comport with convention. One knows that 5 divides evenly into 365, but would be less conventional. In this case the 366th resampled day could be neglected or fudged (if not already handled via the notte bene above) - it occurs once per four years and being a resampled day has 1/7 the weight of an excess week, and half the weight of two dry lumps. One also knows that 6 divides evenly into 366, though that would seem benefit the less frequent leap years and dirty the non-leaps.

edited Mar 4A 365-day year contains 52.14286 weeks, having one small dry lump in the batter for one day excess, whereas a 366-day year has 52.28571 weeks and two dry lumps. Therefore even the 52-week years contain errors that just didn’t rise to the level of adding an additional week to the annum.

(edit: Also Each year begins on a different day that splits the werk partially to the prior year with remainder to current year. I presume there is an accepted procedure for declaring a 53-week year or it simply might be a 53rd batch if reports is recorded by year-end. In any case, it seems to me that all years have some days of error due to the residual days.)

One thought is whether employment of a resampling algorithm may provide benefit. A first approach is to resample each 53-week year to a 52-week year (or all 52-week yeara to a 53).

A second approach is to resample each year to days, then choose a convenient integer for a unit of time with which to do comparisons (more on that further down).

There are theoreticallly four resamplings to consider - 52 week data for a 365-day year, 52-week data for a 366-day year, and two cases for 53-week years. Resampling ratios: 52/365, 52/366, 53/365, 53/366.

N.B. the 366-day years could be resampled to a 365 day year as a way of neglecting the extra day in each leap year.

The result would be a dataset of resampled days for each year. Next, regrouping is in order to make comparisons tastier.

Choosing 7 re-sampled days for a comparison ’week’ would seem to comport with convention. One knows that 5 divides evenly into 365, but would be less conventional. In this case the 366th resampled day could be neglected or fudged (if not already handled via the notte bene above) - it occurs once per four years and being a resampled day has 1/7 the weight of an excess week, and half the weight of two dry lumps. One also knows that 6 divides evenly into 366, though that would seem benefit the less frequent leap years and dirty the non-leaps.