A112404 a(n) = (Product_{e_i=0..1} (Product_{i=1..n} p_i^e_i + Product_{i=1..n} p_i^(1-e_i)))^(1/2) where p_i is the i-th prime.

3, 35, 75361, 105640931881921, 368107372881122974005026861194791580321, 1068920105772796102633531337368359482127315843763564268088796774223747755119986736765386063992951681

Offset: 1

Author

Yasutoshi Kohmoto

Keywords

• nonn

Comments

This is a "Proof of existence of infinitely many primes" sequence. Proof. Let N = (Product_{e_i=0..1} (Product_{i=1..n} p_i^e_i + Product_{i=1..n} p_i^(1-e_i)))^(1/2). Suppose there are only a finite number of primes p_i, 1 <= i <= n. If N is prime, then for all i, N != p_i because, for all i, p_i < N. If N is composite, then it must have a prime divisor p which is different from primes p_i because, for all i, N !== 0 (mod p_i).

The numbers of decimal digits of a(n) are 1, 2, 5, 15, 39, 100, 246, 590, 1387, 3215, 7321, 16507, 36823, 81305, 178212, 388495, 842638, 1816984, ..., . - Robert G. Wilson v

The numbers of prime factors of a(n) are 1, 2, 4, 8, 16, 33, 69, 136, 280, 566, 1107, ..., . - Robert G. Wilson v

Examples

			a(3) = ((1 + p_1*p_2*p_3)*(p_3 + p_1*p_2)*(p_2 + p_1*p_3)*(p_2*p_3 + p_1)*(p_1 + p_2*p_3)*(p_1*p_3 + p_2)*(p_1*p_2 + p_3)*(p_1*p_2*p_3 + 1))^(1/2)
= (1 + p_1*p_2*p_3)*(p_3 + p_1*p_2)*(p_2 + p_1*p_3)*(p_2*p_3 + p_1) = 31*11*13*17.
		

Crossrefs

Cf. A111392.

Programs

Mathematica

f[n_] := Block[{a = 1, p = Prime@Range@n, k = 0, lmt = 2^(n - 1)}, While[k < lmt, e = IntegerDigits[k, 2, n]; a = a*(Times @@ (p^e) + Times @@ (p^(1 - e))); k++ ]; a]; Array[f, 7] (* Robert G. Wilson v *)

Extensions

Edited by Robert G. Wilson v, Dec 10 2005