A284761 a(n) = gcd(A279513(n), A279513(n+1)).

1, 1, 1, 1, 1, 1, 1, 6, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 4, 2, 1, 1, 1, 1, 1, 3

Offset

1,8

Comments

Two consecutive numbers, say n and n+1, cannot share a prime factor (gcd(n, n+1)=1). However, their prime tower factorizations can share some prime numbers; this is the case iff a(n)>1 (see A182318 for the definition of the prime tower factorization of a number).

If p is prime, then a(p-1) = a(p) = 1.

If p is an odd prime, then a(p^2) = 2.

This sequence contains a multiple of p for any prime p:

- let m = A074792(p)^p-1,

- m is a multiple of p, hence p divides A279513(m),

- m+1 = A074792(p)^p, hence p divides A279513(m+1),

- hence p divides gcd(A279513(m), A279513(m+1)) = a(m).

This sequence contains infinitely many distinct values; see A284821 for these distinct values in order of appearance, and A284822 for the corresponding indexes.

Links

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Rémy Sigrist, Illustration of the first terms

Example

a(8) = gcd(A279513(8), A279513(9)) = gcd(A279513(2^3), A279513(3^2)) = gcd(2*3, 3*2) = 6.

Crossrefs

Cf. A074792, A182318, A284821, A284822.

Sequence in context: A171542 A335587 A320302 * A021165 A165061 A306764

Adjacent sequences: A284758 A284759 A284760 * A284762 A284763 A284764

Keyword

nonn

Author

Rémy Sigrist, Apr 02 2017

Status

approved