

A113270


a(n) = sqrt(Product_{k=1..2^n} (Product_{i=1..n} p_i^e_{k,i} + Product_{i=1..n} p_i^(1e_{k,i}))) * Sum_{i=1..n} ((1/p_i)*Product_{k=1..n} p_k) where p_i is the ith prime and e_{k,i} is a vector of length n that runs through all combinations of {0,1}.


1




OFFSET

1,1


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^(1e_i)))^(1/2) * (Sum_{i=1..n} (1/p_i*Product_{k=1..n} p_k)). 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).


LINKS

Amiram Eldar, Table of n, a(n) for n = 1..8


EXAMPLE

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) * (p_2*p_3 + p_1*p_3 + p_1*p_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) * (p_2*p_3 + p_1*p_3 + p_1*p_2)
= 31*11*13*17*31.


MATHEMATICA

a[n_] := Module[{e = Tuples[{0, 1}, n]}, (Product[Product[Prime[i]^e[[j]][[i]], {i, 1, n}] + Product[Prime[i]^(1  e[[j]][[i]]), {i, 1, n}], {j, 1, 2^n}])^(1/2) ]*Sum[1/Prime[i], {i, 1, n}]* Product[Prime[i], {i, 1, n}]; Array[a, 6] (* Amiram Eldar, Nov 23 2018 *)


CROSSREFS

Cf. A111392.
Sequence in context: A300105 A125711 A268949 * A091324 A198446 A216967
Adjacent sequences: A113267 A113268 A113269 * A113271 A113272 A113273


KEYWORD

nonn


AUTHOR

Yasutoshi Kohmoto, Jan 07 2006


EXTENSIONS

Name clarified by and a(5) from Amiram Eldar, Nov 23 2018


STATUS

approved



