A083544 a(n) = maximal value of the sum of Mobius function values over a block of n consecutive natural numbers.

1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 9, 9, 10, 11, 12, 12, 12, 13, 14, 14, 15, 16, 17, 17, 18, 18, 19, 19, 20, 21, 22, 22, 23, 24, 24, 24, 25, 26, 27, 27, 28, 29, 30, 30, 31, 32, 33, 33, 34, 34, 35, 35, 36, 37, 38, 38, 39, 39, 40, 40, 41, 42, 43, 43, 44, 45, 45, 45, 46, 47, 48, 48, 49, 49, 50, 50

Offset: 1

Author

Yuval Dekel (dekelyuval(AT)hotmail.com), Jun 10 2003

Keywords

• nonn

Comments

Comment from Hugh Montgomery (hlm(AT)umich.edu): I do not recall having seen literature on this question. If p is a prime, p < sqrt(k), then there will be a multiple of p^2 in the block and such a number will then contribute 0. Let Q(M, k) denote the numbers of integers between M+1 and M+k (inclusive) that are not divisible by the square of any prime <= sqrt(k). By the sieve of Eratosthenes-Legendre, Q(M,k) = k/zeta(2) +O(sqrt(k)), uniformly in M. Let Q^+(k) = max_M Q(M,k). I expect that the sum of mu(n) over n = M+1..M+k can be as large as Q^+(k) and as small as -Q^+(k). Indeed, I expect that this could be shown to follow from the prime k-tuple conjecture.

The maximum first appears at A225420(n). - T. D. Noe, May 07 2013

Crossrefs

Cf. A008683, A063838, A082967, A225420.

Formula

a(n) = max sum m=i...(i+n-1) Mobius(m) over i>=1.

Extensions

Offset corrected by Eric M. Schmidt, May 07 2013

More terms from Don Reble, Apr 21 2021