A251600 Least k such that prime(k) + prime(k+1) contains n prime divisors (with multiplicity), otherwise 0.

1, 0, 2, 5, 16, 20, 18, 43, 162, 190, 532, 916, 564, 3314, 3908, 10499, 30789, 53828, 153384, 62946, 278737, 364195, 629686, 3768344, 7827416, 9496221, 23159959, 184328920, 68035462, 92566977, 457932094, 370110663, 648634305, 4032924162, 7841376455

Offset: 1

Author

Michel Lagneau, Dec 05 2014

Keywords

• nonn

Comments

If p and q are two consecutive odd primes, then p + q is the product of at least three primes (not necessarily distinct) because p + q = 2*(p + q)/2 => (p + q)/2 is a composite integer between two consecutive primes p and q that is the product of at least two prime numbers. Thus 2*(p + q)/2 has at least three prime factors => a(1) = 1 because prime(1) is even => prime(1) + prime(2) = 5 is prime and a(2) = 0, probably the only 0 of the sequence.

Examples

			a(5) = 16 because prime(16) + prime(17) = 53 + 59 = 112 = 7*2^4 with 5 prime divisors.
		

Links

• Giovanni Resta, Table of n, a(n) for n = 1..45

Crossrefs

Cf. A098037, A001222.

Programs

Mathematica

A251600 = {1, 0}; Do[k = 1; While[PrimeOmega[Prime[k] + Prime[k + 1]] != n, k++]; AppendTo[A251600, k], {n, 3, 10}]; A251600

Extensions

a(28)-a(33) from Daniel Suteu, Nov 18 2018

a(34)-a(35) from Giovanni Resta, Nov 19 2018