Solutions
Solution 1 (Click to Show/Hide)
// Initalize prime number list with seive
const NUM_PRIMES = 1000000;
const PRIMES = [2];
const PRIME_SEIVE = Array(Math.floor((NUM_PRIMES-1)/2)).fill(true);
(function initPrimes(num) {
const upper = Math.floor((num - 1) / 2);
const sqrtUpper = Math.floor((Math.sqrt(num) - 1) / 2);
for (let i = 0; i <= sqrtUpper; i++) {
if (PRIME_SEIVE[i]) {
// Mark value in PRIMES array
const prime = 2 * i + 3;
PRIMES.push(prime);
// Mark all multiples of this number as false (not prime)
const primeSqaredIndex = 2 * i ** 2 + 6 * i + 3;
for (let j = primeSqaredIndex; j < upper; j += prime)
PRIME_SEIVE[j] = false;
}
}
for (let i = sqrtUpper + 1; i < upper; i++) {
if (PRIME_SEIVE[i])
PRIMES.push(2 * i + 3);
}
})(NUM_PRIMES);
function isPrime(num) {
if (num === 2)
return true;
else if (num % 2 === 0)
return false
else
return PRIME_SEIVE[(num - 3) / 2];
}
function consecutivePrimeSum(limit) {
// Initalize for longest sum < 100
let bestPrime = 41;
let bestI = 0;
let bestJ = 5;
// Find longest sum < limit
let sumOfCurrRange = 41;
let i = 0, j = 5;
// -- Loop while current some starting at i is < limit
while (sumOfCurrRange < limit) {
let currSum = sumOfCurrRange;
// -- Loop while pushing j towards end of PRIMES list
// keeping sum under limit
while (currSum < limit) {
if (isPrime(currSum)) {
bestPrime = sumOfCurrRange = currSum;
bestI = i;
bestJ = j;
}
// -- Increment inner loop
j++;
currSum += PRIMES[j];
}
// -- Increment outer loop
i++;
j = i + (bestJ - bestI);
sumOfCurrRange -= PRIMES[i - 1];
sumOfCurrRange += PRIMES[j];
}
// Return
return bestPrime;
}
consecutivePrimeSum(1000000);