uint8_t rtc_DOW(uint8_t y, uint8_t m, uint8_t d)
{
// Added 1 to Jan, Feb.
// Subtracted 1 from each instead of adding 6 in calc below.
static const uint8_t mTable[] = {0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4};

// Year is 0 – 99, rep years 2000 – 2099
// Month starts at 0.
// Day starts at 1.

// Subtract 1 in calc for Jan, Feb, only in leap years.
// Subtracting 1 from year has the effect of subtracting 2 in leap years, subtracting 1 otherwise.
// Adding 1 for Jan, Feb in Month Table so calc ends up subtracting 1 for Jan, Feb, only in leap years.
// All of this complication to avoid the check if it is a leap year.
if (m > 2)) % 7;

} /* end */

uint8_t rtc_DayOfWeek(uint8_t year, uint8_t month, uint8_t day)
{
//static const uint8_t month_offset_table[] = {0, 3, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5}; // Typical table.

// Added 1 to Jan, Feb. Subtracted 1 from each instead of adding 6 in calc below. */
static const uint8_t month_offset_table[] = {0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4};

uint8_t day_of_week;

// Year is 0 – 99, representing years 2000 – 2099

// Month starts at 0.
month -= 1;

// Day starts at 1.

// Subtract 1 in calc for Jan, Feb, only in leap years.
// Subtracting 1 from year has the effect of subtracting 2 in leap years, subtracting 1 otherwise.
// Adding 1 for Jan, Feb in Month Table so calc ends up subtracting 1 for Jan, Feb, only in leap years.
// All of this complication to avoid the check if it is a leap year.
if (month > 2)) % 7;

// Sunday (0) …
return day_of_week;

} /* end rtc_DayOfWeek() */

Enjoy.

According to the algorithm:
1. From the Centuries Table, 2000 corresponds to 6.
2. The last two digits if the year are simply 0.
3. zero mod 4 is 0.
4. From the Month Table, for a leap year (2000) January’s code is 6.
5. the sum of 6 + 0 + 0 + 6 is 12.
6. 12 mod 7 is 5.
7. The implied day is Friday.

But the actual day is Saturday!

The algorithm doesn’t work, even for your example walk-through.
January 31, 1883 was a Wednesday, not a Sunday.