%I #70 Sep 06 2014 00:42:05
%S 4,1,4,6,8,5,4,10,6,7,5,4,4,8,9,6,8,5,4,6,7,5,4,4,4,9,8,10,6,7,5,4,4,
%T 5,4,14,3,3,1,1,6,10,9,8,8,6,6,8,8,6,6,10,5,4,5,3,4,1,4,4,4,5,3,26,1,
%U 4,24,10,9,8,17,6,16,13,12,15,10,9,8,10,6,7
%N a(n) is the smallest k such that the sum of n consecutive values M(k) + M(k+1) + ... + M(k+n-1) is zero, where M(m) is the Moebius (or Möbius) function (A008683).
%H Michel Lagneau, <a href="/A244097/b244097.txt">Table of n, a(n) for n = 1..1000</a>
%e a(1)= 4 => M(4) = 0;
%e a(2)= 1 => M(1)+ M(2) = 1-1 = 0;
%e a(3)= 4 => M(4)+ M(5)+ M(6) = 0-1+1 = 0;
%e a(4)= 6 => M(6)+ M(7)+ M(8) + M(9) = 1-1+0+0 = 0;
%e a(5)= 8 => M(8)+ M(9)+ M(10)+ M(11)+ M(12) = 0+0+1-1+0 = 0.
%t Table[k=1;While[Sum[MoebiusMu[k+i],{i,0,n-1}]!=0,k++];k,{n,1,100}]
%o (PARI) a(n) = my(k = 1); while(sum(j=k, n+k-1, moebius(j)) != 0, k++); k; \\ _Michel Marcus_, Aug 30 2014
%Y Cf. A008683, A171910.
%K nonn
%O 1,1
%A _Michel Lagneau_, Aug 30 2014
|