cp's OEIS Frontend

A383115 a(n) is the number of values m that satisfy floor(sqrt(m)) = n and A381054(m) = 1.

%I A383115 #26 May 14 2025 16:28:25
%S A383115 0,3,4,2,0,2,6,8,5,3,0,3,7,10,11,9,7,2,0,2,8,12,13,15,12,10,8,3,0,3,8,
%T A383115 11,16,19,18,16,13,11,7,3,0,2,7,11,16,18,24,22,21,17,12,12,8,3,0,3,7,
%U A383115 13,15,18,25,24,28,22,21,18,13,11,7,3,0,3,6,11,16,20,23,27,29,30
%N A383115 a(n) is the number of values m that satisfy floor(sqrt(m)) = n and A381054(m) = 1.
%C A383115 The sequence is quasiperiodic with quasiperiod 2*n + 1.
%C A383115 Conjecture: Sum_{n=1..N} a(n) ~ 4*N^(3/2)/3 for N large enough.
%C A383115 Conjecture: Sum_{n=1..N} a(n) ~ 2*Sum_{n=1..N} A382306(n) for N large enough.
%F A383115 a(n) = |{m: n^2 <= m <= n^2+2*n and C(m)=1}| where C(m) = floor(sqrt(A033677(m))) - floor(sqrt(A033676(m))).
%e A383115 a(1) = 0 because C(1) = 5, C(2) = 4, C(3) = 4.
%e A383115 a(2) = 3 because C(4)..C(8) = 2,1,2,1,1 and only three arguments satisfy C(m) = 1.
%e A383115 a(3) = 4 because C(9)..C(15) = 2,1,4,1,4,1,1 and only four arguments satisfies C(m) = 1.
%e A383115 a(4) = 2 because C(16)..C(24) = 5,9,1,9,3,1,2,9,3 and only two arguments satisfying C(m) = 0.
%o A383115 (PARI) d(n) = if(n<2, 1, my(d=divisors(n)); d[(length(d)+1)\2]); \\ A033676
%o A383115 f(n) = my(k=1); while (sqrtint(n*k/d(n*k)) - sqrtint(d(n*k)) != 1, k++); k; \\ A381054
%o A383115 a(n) = #select(x->f(x)==1, [n^2..n^2+2*n]);
%Y A383115 Cf. A033676, A033677, A382286, A381054, A382306.
%K A383115 nonn
%O A383115 1,2
%A A383115 _Hassan Baloui_, Apr 16 2025