A093429 Number of distinct prime factors of (prime(1)*...*prime(n))+(prime(n+1)*...*prime(2n)), where prime(n) is the n-th prime.

1, 1, 1, 1, 2, 2, 2, 3, 2, 3, 2, 2, 3, 2, 2, 4, 3, 2, 6, 3, 4, 4, 3, 1, 1, 3, 3, 3, 3, 2, 4, 3, 3, 3, 3, 5, 4, 2, 3, 3, 5, 3, 7, 4, 1, 4, 3, 4, 3, 6, 2, 4, 3, 3

Offset: 1

Author

Jason Earls, May 12 2004

Keywords

• nonn
• more

Comments

Prime for n = 1, 2, 3, 4, 24, 25, 45, 59 and no more for n < 100 (A329532).

Examples

			a(31)=4 because 509102378439545188849067644696085192959414195658632710736111053092210207
= 3711597629 * 238694867020723 * 226814268663739929299 * 2533557617597929944840907379.
		

Links

• Dario Alejandro Alpern, Factorization using the Elliptic Curve Method.
• P. Samidoost, Primenumbers group posting.
• Cashogor, Payam Samidoost, David Cleaver, Jens Kruse Andersen, Creating Primes, digest of 9 messages in primeforms Yahoo group, May 12, 2004. [Cached copy]

Crossrefs

Cf. A001221, A002110, A093433, A329532.

Programs

Mathematica

PrimeFactors[n_] := Flatten[ Table[ # [[1]], {1} ] & /@ FactorInteger[n]]; f[n_] := Length[ PrimeFactors[ Product[Prime[i], {i, n}] + Product[Prime[i + n], {i, n}]]]; Table[ f[n], {n, 20}]

PARI

a(n) = omega(prod(k=1, n, prime(k)) + prod(k=n+1, 2*n, prime(k))); \\ Daniel Suteu, Nov 26 2019

Formula

a(n) = A001221(A002110(n) + A002110(2*n) / A002110(n)). - Daniel Suteu, Nov 26 2019

Extensions

a(40)-a(48) from Robert G. Wilson v, May 27 2004

a(49)-a(54) from Daniel Suteu, Nov 26 2019