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a(n) = maximal value of the sum of Mobius function values over a block of n consecutive natural numbers.
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%I #13 Apr 21 2021 10:23:26

%S 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,

%T 19,20,21,22,22,23,24,24,24,25,26,27,27,28,29,30,30,31,32,33,33,34,34,

%U 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

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

%C 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.

%C The maximum first appears at A225420(n). - _T. D. Noe_, May 07 2013

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

%Y Cf. A008683, A063838, A082967, A225420.

%K nonn

%O 1,2

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

%E Offset corrected by _Eric M. Schmidt_, May 07 2013

%E More terms from _Don Reble_, Apr 21 2021