A089497 mu(prime(n)+1) - mu(prime(n)-1), where mu is the Moebius function.

1, 1, -1, -1, 1, 0, 0, -1, -1, 1, 1, -1, 1, -1, 0, -1, 1, 1, 1, 1, 1, -1, 0, 0, -1, 1, -1, -1, -1, 0, 1, -1, 1, 0, 0, 1, 0, -1, -1, -1, -1, 1, 1, 0, 0, -1, 1, -1, -1, 0, 1, 0, 0, -1, -1, 0, 0, 1, -1, 1, 0, 0, 1, 1, -1, -1, 0, -1, 0, -1, -1, 1, -1, 0, -1, 1, 1, -1, -1, 1, 1, 1, -1, 1, 1, 0, 1, 1, -1, -1, -1, 0, 0, 1, -1, 1, 0, 0, 1, -1, 0, -1, 1, -1, 0, -1, 0

Offset: 2

Author

T. D. Noe, Nov 04 2003

Keywords

• sign

Comments

This difference is always -1, 0 or 1 because for odd prime p, both p-1 and p+1 cannot be squarefree; one of them will be divisible by 4. This also implies that terms in this sequence are zero only for primes p such that mu(p-1) = mu(p+1) = 0, which is A075432.

Links

• Eric Weisstein's World of Mathematics, Moebius Function
• Eric Weisstein's World of Mathematics, Legendre Symbol

Crossrefs

Cf. A089451 (mu(p-1) for prime p), A089495 (mu(p+1) for prime p), A089496 (mu(p+1)+mu(p-1) for prime p).

Programs

Mathematica

Table[MoebiusMu[Prime[n]+1] - MoebiusMu[Prime[n]-1], {n, 2, 150}]
MoebiusMu[#+1]-MoebiusMu[#-1]&/@Prime[Range[2,110]] (* Harvey P. Dale, Sep 16 2018 *)

Formula

Let p = prime(n), then a(n) = (-1/p) mu(p+(-1/p)), where (-1/p) is the Legendre symbol, A070750. (Pieter Moree)