A214874 Starting with Fibonacci(0), the sum of a(n) successive Fibonacci numbers is prime.

3, 1, 1, 1, 5, 1, 5, 1, 5, 1, 11, 13, 131, 31, 65, 49, 47, 13, 2231, 389, 5269, 72211, 12587, 51193

Offset: 1

Author

Alex Ratushnyak, Jul 28 2012

Keywords

• nonn
• hard
• more

Comments

a(22), if it exists, is bigger than 60300.

The sequence with corresponding primes begins: 2, 2, 3, 5, 131, 89, 2351, 1597, 42187, 28657, 14855327, 7763811697. The prime corresponding to a(21) = 5269 has 1729 decimal digits.

Examples

			0+1+1 = 2 is prime, three summands,
2 is prime,
3 is prime,
5 is prime,
8+13+21+34+55 = 131 is prime, five summands,
89 is prime,
144+233+377+610+987 = 2351 is prime, five summands,
1597 is prime.
		

Crossrefs

Cf. A000040, A000045, A073684, A214878.

Programs

Java

import static java.lang.System.out;
import java.math.BigInteger;
public class A214874 {
  public static void main (String[] args) {
    long i, n=0;
    BigInteger prpr = BigInteger.ZERO;
    BigInteger prev = BigInteger.ONE, curr;
    while (true) {
      BigInteger bsum = BigInteger.ZERO;
      for (i=n; ; ++i) {
        bsum = bsum.add(prpr);
        curr = prev.add(prpr);
        prpr = prev;
        prev = curr;
        if (bsum.isProbablePrime(2)) {
                if (bsum.isProbablePrime(80)) break;
                out.printf("(%d)",i);
        }
      }
      out.printf("%d, ", i+1-n);
      n=i+1;
    }
  }
}

Extensions

a(22)-a(24) from Michael S. Branicky, Nov 21 2024