A083552 Quotient when LCM of 2 consecutive prime differences is divided by GCD of the same two differences.

2, 1, 2, 2, 2, 2, 2, 6, 3, 3, 6, 2, 2, 6, 1, 3, 3, 6, 2, 3, 6, 6, 12, 2, 2, 2, 2, 2, 14, 14, 6, 3, 5, 5, 3, 1, 6, 6, 1, 3, 5, 5, 2, 2, 6, 1, 3, 2, 2, 6, 3, 5, 15, 1, 1, 3, 3, 6, 2, 5, 35, 14, 2, 2, 14, 21, 15, 5, 2, 6, 12, 12, 1, 6, 6, 12, 2, 2, 20, 5, 5, 5, 3, 6, 6, 12, 2, 2, 2, 3, 6, 2, 2, 2, 6, 2, 6, 9

Offset: 1

Author

Labos Elemer, May 22 2003

Keywords

• nonn

Comments

Conjecture: Every positive integer appears infinitely many times in this sequence. Example: a(834) = a(909) = ... = a(9901) = ... = 4. - Jerzy R Borysowicz, Dec 22 2018

All terms of this sequence are integers because gcd(r,s) divides lcm(r,s) for any r and s. - Jerzy R Borysowicz, Jan 05 2019

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001223, A083538-A083555, A057467.

Programs

Mathematica

f[x_] := Prime[x+1]-Prime[x]; Table[LCM[f[w+1], f[w]]/GCD[f[w+1], f[w]], {w, 1, 128}]

PARI

a(n) = my(da=prime(n+2)-prime(n+1), db=prime(n+1)-prime(n)); lcm(da, db)/gcd(da, db) \\ Felix Fröhlich, Jan 05 2019

Formula

a(n) = lcm(A001223(n), A001223(n+1))/gcd(A001223(n), A001223(n+1));

a(n) = A083551(n)/A057467(n).