How to calculate if a year has 53 weeks

Question

According to formula https://en.wikipedia.org/wiki/ISO_week_date in the Weeks per year section you should be able to find each year with 53 weeks. I have copied the formula into PHP but it seems to mess up around 2020 returning 52 weeks instead of 53.

function weeks($year){
    $w = 52;
    $p = ($year+($year/4)-($year/100)+($year/400))%7;
    if ($p == 4 || ($p-1) == 3){
        $w++;
    }
    return $w." ".$p;
}

for ($i = 2000; $i <= 2144; $i++) {
    echo $i." (".weeks($i).") | ";
}

From Wikipedia

The following 71 years in a 400-year cycle have 53 weeks (371 days); years not listed have 52 weeks (364 days); add 2000 for current years:

004, 009, 015, 020, 026, 032, 037, 043, 048, 054, 060, 065, 071, 076, 082, 088, 093, 099, 105, 111, 116, 122, 128, 133, 139, 144

I know I can grab the total amount of weeks for a given year using the date() function but I'm just wondering if anyone knows the math since the formula on wikipedia seems to be wrong.

EDIT WORKING CODE

function p($year){
    return ($year+floor($year/4)-floor($year/100)+floor($year/400))%7;
}

function weeks($year){
    $w = 52;
    if (p($year) == 4 || p($year-1) == 3){
        $w++;
    }
    return $w; // returns the number of weeks in that year
}

for ($i = 2000; $i <= 2144; $i++) {
    echo $i." (".weeks($i).") | ";
}

Answer 1

Two problems that I can see:

1. You missed the floor notation in the formula for p. In PHP that should be + int($year / 4) - int($year / 100) etc. not + ($year / 4) - ($year / 100) etc.

2. p is a function, and the overall formula uses p(year) and p(year - 1). Your $p is equal to p(year), but $p - 1 is not equal to p(year - 1) (note that your second condition, ($p - 1) == 3 is just the same as your first condition, $p == 4 — this should make it clear that that's not what was intended). The easiest fix is to write it as a function in PHP also.