A081221 Number of consecutive numbers >= n having at least one square divisor > 1.

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

Offset: 1

Author

Reinhard Zumkeller, Mar 10 2003

Keywords

• nonn

Comments

The first time terms 0..7 occur is at n = 1, 4, 8, 48, 242, 844, 22020, 217070. - Antti Karttunen, Sep 22 2017

Examples

			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
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537

Crossrefs

Cf. A005117, A008683, A045882, A051681, A068781, A067535, A080733.

Programs

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 *)

PARI

A081221(n) = { my(k=0); while(!issquarefree(n+k),k++); k; }; \\ Antti Karttunen, Sep 22 2017

Python

from itertools import count
from sympy import factorint
def A081221(n): return next(m for m in count(0) if max(factorint(n+m).values(),default=0)<=1) # Chai Wah Wu, Dec 04 2024

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.

a(A068781(n)) > 0.

a(n) = A067535(n) - n. - Amiram Eldar, Oct 10 2023