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

%I A081221 #30 Dec 06 2024 11:09:41
%S A081221 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,
%T A081221 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,
%U A081221 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
%N A081221 Number of consecutive numbers >= n having at least one square divisor > 1.
%C A081221 The first time terms 0..7 occur is at n = 1, 4, 8, 48, 242, 844, 22020, 217070. - _Antti Karttunen_, Sep 22 2017
%H A081221 Antti Karttunen, <a href="/A081221/b081221.txt">Table of n, a(n) for n = 1..65537</a>
%F A081221 mu(k) = 0 for n <= k < n+a(n) and mu(n+a(n)) <> 0, where mu = A008683 (Moebius function).
%F A081221 a(n)*mu(n) = 0.
%F A081221 a(A068781(n)) > 0.
%F A081221 a(n) = A067535(n) - n. - _Amiram Eldar_, Oct 10 2023
%e A081221 For n = 3, 3 is a squarefree number, thus a(3) = 0.
%e A081221 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
%t A081221 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 *)
%o A081221 (PARI) A081221(n) = { my(k=0); while(!issquarefree(n+k),k++); k; }; \\ _Antti Karttunen_, Sep 22 2017
%o A081221 (Python)
%o A081221 from itertools import count
%o A081221 from sympy import factorint
%o A081221 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
%Y A081221 Cf. A005117, A008683, A045882, A051681, A068781, A067535, A080733.
%K A081221 nonn
%O A081221 1,8
%A A081221 _Reinhard Zumkeller_, Mar 10 2003