In Javascript you can use this:

getWeekDay = function (year) {
  var d = new Date(); 
  d.setFullYear(year,0,1);
  return d.getDay()+1;
};

document.write(getWeekDay(2011));

Result is 1..7, as requested.

As reported also by Wikipedia, in 1990 Michael Keith and Tom Craver published an expression to minimise the number of keystrokes needed to enter a self-contained function for converting a Gregorian date into a numerical day of the week.

The expression does preserve neither y nor d, and returns a zero-based index representing the day, starting with Sunday, i.e. if the day is Monday the expression returns 1.

A code example which uses the expression follows:

int d    = 15   ; //Day     1-31
int m    = 5    ; //Month   1-12`
int y    = 2013 ; //Year    2013` 

int weekday  = (d += m < 3 ? y-- : y - 2, 23*m/9 + d + 4 + y/4- y/100 + y/400)%7;  

The expression uses the comma operator, as discussed in this answer.

Enjoy! ;-)

mktime() uses many of the members in struct tm, so calling mktime() without setting those members results in undefined behavior (UB).

// struct tm time;
struct tm time = { 0 };  // set all fields to 0.
...    
time.tm_year = year - 1900;
time.tm_mon = month - 1;
time.tm_mday = day;
time_t rightTime = mktime(&time);

After mktime(), the day-of-the-week member is set.

int days_since_Sunday = time.tm_wday; // 0 - 6

... how to find out the first day of that given week, refresh all the information and then print it out for user. ... print the whole week to user from Monday till Sunday.

To find the days since Monday, subtract 1, but use modulo math and avoid negative numbers. % matches the classic functionality of modulo with positive arguments, but differs with negative ones.

#define DaysPerWeek 7
int days_since_Monday = (days_since_Sunday + DaysPerWeek - 1) % DaysPerWeek;

To print the week, start with the given date and subtract the offset:

// Move date back to Monday
time.tm_mday -= days_since_Monday;

for (int d=0; d<DaysPerWeek; d++) {
  mktime(&time); // adjust for end of month, year issues.
  strftime(buffer, sizeof buffer, "%A %d. %B %Y\n", &time);
  printf("%s", buffer);
  time.tm_mday++; // advance to tomorrow
}

Note: using time as in struct tm time is OK, but is a bit confusing given the standard function time(). Consider using a different name. Perhaps struct tm timestamp_ymd.

Robust code would add error checking:

if (mktime(&time) == (time_t)-1) Handle_Error();

Note: Not tested, but given the current year, this should do it:

const char *months[12]={"Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep",
                        "Oct","Nov","dec","Jan"};
/* Start with January 1st of the current year */
struct tm curYear={
  0, // secs
  0, // mins
  0, // hours
  1, // Day of month
  0, // Month (Jan)
  year,
  0, // wday
  0, // yday
  0}; // isdst

/* Offset the number of weeks specified */
time_t secsSinceEpoch=mktime(&curYear)+
                      weekNum*86400*7; /* Shift by number of weeks */
struct tm *candidateDate=gmtime(&secsSinceEpoch);

/* If the candidate date is not a Monday, shift it so that it is */
if (candidateDate->tm_wday!=1)
{
  secsSinceEpoch+=(86400*(candidateDate->tm_wday-1)); 
  candidateDate=gmtime(&secsSinceEpoch);
}

printf("Mon %s %d",months[candidateDate->tm_mon],candidateDate->tm_mday\n");

You may have to adjust the formulas in this code depending on what exactly you mean by week 43 of a given year or to conform with ISO-8601, for example. However, this should present you with good boiler plate code to get started. You may also want to parameterize the day of the week, so that it is not hard coded.

Also, if you want, you can avoid the months array and having to format the time, by truncating the result of the ctime function, which just so happens to display more than you asked for. You would pass to it a pointer to the secsSinceEpoch value and truncate its output to just display the day of the week, the day of the month and the abbreviation of the months name.

Specifying the first day of the first week manually

You can use pgfcalendar to store the first day of your first week in a LaTeX counter and use that later as the base for an offset.

With \setStartDate you set the first day of your first week and the \getDateOfWeek{<week>}{<offset>} will define \Year, \Month, \Day with the date that is in the <week>th week with an additional <offset>.

For example:

\setStartDate{2022-10-10}

sets the first week to start on October 10th, 2022.

With

\getDateOfWeek{1}{4}

you'll get October 14th and with

\getDateOfWeek{2}{0}

you'll get October 17th in \Month and \Day.

Code

\documentclass{article}
\usepackage{pgfcalendar}
\newcounter{myStartDate}
\makeatletter
\newcommand*{\setStartDate}[1]{% for week 1
  \begingroup
    \pgfcalendardatetojulian{#1}{\@tempcnta}%
    \setcounter{myStartDate}{\@tempcnta}%
  \endgroup
}
\newcommand*{\getDateOfWeek}[2]{%
  \begingroup
    \@tempcnta=\inteval{(#1-1)*7+#2}\relax
    \advance\@tempcnta by \c@myStartDate
    \edef\@temp{\endgroup\noexpand\pgfcalendarjuliantodate{\the\@tempcnta}}%
    \@temp{\Year}{\Month}{\Day}%
}
\makeatother
\usepackage{pgffor}
\begin{document}
\setStartDate{2022-01-03}
\foreach \week in {1,...,52}{
  Week \week\ goes from \getDateOfWeek{\week}{0}\Month/\Day\ to
                        \getDateOfWeek{\week}{4}\Month/\Day.\par}
\end{document}

Output

Using the actual week numbers of the years.

With the pgfcalendar-ext package of my tikz-ext package which incorporates another answer of mine and uses the week numbering according to ISO 8601 which mainly means that a new week starts on Monday.

Here's a solution with

\getTwoDaysOfWeek[<year>]{<week>}{<offset>}
  {<year>}{<month>}{<day>}{<offset year>}{<offset month>}{<offset day>}

which defines six macros (the argments on the second line) for you which contain

• the first day of <week> of <year> and
• the day that's <offset> after the first day of <week> of <year>.

The <year> argument is optionally, the default is the current year.

Code

\documentclass{article}
\usepackage{pgfcalendar-ext}
\newcommand*{\stripZero}[1]{\if0#1\else#1\fi}
\makeatletter
\newcommand*{\getTwoDaysOfWeek}[9][\the\year]{%
  % #1 = year (defaults to current year)
  % #2 = week number, #3 = offset
  % #4/#5/#6 = start year/month/day
  % #7/#8/#9 = end year/month/day
  \begingroup
    \pgfcalendardatetojulian{#1-01-01}{\@tempcnta}%
    \pgfcalendarjuliantoweekday{\@tempcnta}{\@tempcntb}%
    \edef\@dateStartWeek{\the\numexpr\@tempcnta-\@tempcntb+#2*7-7\relax}%
    \pgfcalendarjulianyeartoweek{\@tempcnta}{#1}{\@tempcntb}%
    \ifnum\@tempcntb>1
      % This is the last week of the previous year
      \edef\@dateStartWeek{\the\numexpr\@dateStartWeek+7\relax}%
    \fi
    \edef\@temp{\endgroup%
      \noexpand\pgfcalendarjuliantodate
        {\@dateStartWeek}{\noexpand#4}{\noexpand#5}{\noexpand#6}%
      \noexpand\pgfcalendarjuliantodate
        {\the\numexpr\@dateStartWeek+#3\relax}{\noexpand#7}{\noexpand#8}{\noexpand#9}}%
  \@temp
}
\makeatother
\newcommand*{\weekCell}[2][\the\year]{%
  \begin{tabular}[t]{@{}l@{}}Week #2:\\
    \getTwoDaysOfWeek[#1]{#2}{4}{\y}{\m}{\d}{\Y}{\M}{\D}%
    \stripZero\m/\stripZero\d--\stripZero\M/\stripZero\D
  \end{tabular}}

\usepackage{pgffor}
\usepackage{booktabs}
\begin{document}
\begin{tabular}{lll}
\toprule
Time & Content & Note \\\midrule
\weekCell{1} & Content 1 & Note 1 \\
\weekCell{2} & Content 2 & Note 2 \\
\dots & \dots & \dots \\\bottomrule
\end{tabular}
\end{document}

Output