Find days of the year with a certain sum of digits

findDays.js

const nthDay = (n, year) => {
  const firstDay = new Date(`${year}`);
  const newDay = new Date(firstDay.getTime());
  newDay.setDate(firstDay.getDate() + n);
  
  return newDay;
};

const isLeapYear = (year) =>
  ((year % 4 === 0) && (year % 100 !== 0)) || (year % 400 === 0);

const getAllDays = (year) => {
  const numberOfDays = isLeapYear(year) ? 366 : 365;
  
  return new Array(numberOfDays)
    .fill(0)
    .map((_, index) => index)
    .map(n => nthDay(n, year));
};

const sumOfDigits = (date) =>
  `${date.getUTCFullYear()}${date.getUTCMonth() + 1}${date.getUTCDate()}`
    .split('')
    .map(digit => Number(digit))
    .reduce((acc, cur) => acc + cur, 0);

const findDays = (sum, year) => getAllDays(year).filter(day => sumOfDigits(day) === sum);

To list all dates within the current year with the same sum of digits as the current day:

const addLeadingZero = (n: number) => `${n}`.padStart(2, '0');
const formatDate = (date: Date) => new Intl.DateTimeFormat('fr-FR', { dateStyle: 'short' }).format(date);

const currentDay = new Date();
const foundDays = findDays(sumOfDigits(currentDay), currentDay.getUTCFullYear());
console.log(`${foundDays.length} date(s) found.`);
console.log(foundDays.map(formatDate).join('\n'));

Output:

23 date(s) found.
04/01/2022
13/01/2022
22/01/2022
31/01/2022
03/02/2022
12/02/2022
21/02/2022
02/03/2022
11/03/2022
20/03/2022
02/04/2022
11/04/2022
05/10/2022
14/10/2022
23/10/2022
31/10/2022
03/11/2022
12/11/2022
21/11/2022
30/11/2022
02/12/2022
11/12/2022
20/12/2022