|
|
A081221
|
|
Number of consecutive numbers >= n having at least one square divisor > 1.
|
|
5
|
|
|
0, 0, 0, 1, 0, 0, 0, 2, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 2, 1, 0, 2, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 1, 0, 0, 3, 2, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 2, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 2, 1, 0, 0, 0, 2, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 3, 2, 1, 0, 0, 0, 1, 0
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,8
|
|
COMMENTS
|
The first time terms 0..7 occur is at n = 1, 4, 8, 48, 242, 844, 22020, 217070. - Antti Karttunen, Sep 22 2017
|
|
LINKS
|
|
|
FORMULA
|
mu(k) = 0 for n <= k < n+a(n) and mu(n+a(n)) <> 0, where mu = A008683 (Moebius function).
a(n)*mu(n) = 0.
|
|
EXAMPLE
|
For n = 3, 3 is a squarefree number, thus a(3) = 0.
For n = 48, neither 48 = 2^4 * 3 nor 49 = 7^2, nor 50 = 2^2 * 5 are squarefree, but 51 = 3*17 is, thus a(48) = 3. - Antti Karttunen, Sep 22 2017
|
|
MATHEMATICA
|
Flatten@ Map[If[First@ # == 0, #, Reverse@ Range@ Length@ #] &, SplitBy[Table[DivisorSum[n, 1 &, And[# > 1, IntegerQ@ Sqrt@ #] &], {n, 120}], # > 0 &]] (* Michael De Vlieger, Sep 22 2017 *)
|
|
PROG
|
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|