|
| |
|
|
A083544
|
|
a(n) = maximal value of the sum of Mobius function values over a block of n consecutive natural numbers.
|
|
3
|
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
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.
|
|
|
LINKS
|
|
|
|
FORMULA
|
a(n) = max sum m=i...(i+n-1) Mobius(m) over i>=1.
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
Yuval Dekel (dekelyuval(AT)hotmail.com), Jun 10 2003
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
