A286068 - OEIS

A286068

a(n) = least k such that the prime tower factorizations of k and k+1 both contain the n-th prime.

%I #34 Jan 19 2019 04:14:59

%S 8,8,95,384,10240,57343,1179647,4718592,92274688,8053063679,

%T 32212254720,2611340115967,46179488366591,184717953466368,

%U 3236962232172544,243194379878006783,16717361816799281152,71481133285624512511,4869940435459321626624,82641413450218791239680

%N a(n) = least k such that the prime tower factorizations of k and k+1 both contain the n-th prime.

%C The prime tower factorization of a number is defined in A182318.

%C Two consecutive numbers cannot have a common prime factor; however, their prime tower factorizations can share a prime number.

%C For example, the prime tower factorizations of 8 and 9, that is, 2^3 and 3^2, share the prime numbers 2 and 3.

%C We can also find triples of consecutive numbers whose prime tower factorizations share a prime number:

%C - if n is an odd squarefree number > 1, then the prime tower factorizations of n^2-1, n^2 and n^2+1 share the prime number 2,

%C - the prime tower factorizations of 5344, 5345 and 5346 share the prime number 5.

%C Also, the prime tower factorizations of:

%C - 342, 343, 344 and 345 share the prime number 3,

%C - 99125, 99126, 99127, 99128 and 99129 share the prime number 3,

%C - 72470 ... 72480 share the prime number 2,

%C - 1674274 ... 1674288 share the prime number 2.

%C Are there tuples of more than 15 consecutive numbers with such a property?

%H Rémy Sigrist, <a href="/A286068/a286068.pdf">Illustration of the first terms</a>

%F a(1) = 8.

%F If prime(n) = 4*k+1, then a(n) = 2^(4*k+1)*(2*k+1)-1.

%F If prime(n) = 4*k+3, then a(n) = 2^(4*k+3)*(2*k+1).

%F To prove the formula for n > 1:

%F - we use Fermat's little theorem: 2^p = 2 mod p,

%F - we check that there are no lower values near a multiple of 2^p,

%F - we check that the given value is less than 3^p - 1.

%e See illustration of first terms in Links section.

%o (PARI) a(n) = my (p=prime(n)); if (p==2, return (8), my (k = p\4); if (p % 4 == 1, return (2^p*(2*k+1)-1), return (2^p*(2*k+1))))

%Y Cf. A182318.

%K nonn

%O 1,1

%A _Rémy Sigrist_, Jun 13 2017