cp's OEIS Frontend

A382306 a(n) is the number of values m that satisfy floor(sqrt(m))=n and A382286(m)=1.

%I A382306 #32 Apr 14 2025 18:15:30
%S A382306 3,2,1,3,5,4,2,1,3,5,7,6,4,2,1,3,5,7,9,8,6,4,2,1,3,5,7,9,11,10,8,6,4,
%T A382306 2,1,3,5,7,9,11,13,12,10,8,6,4,2,1,3,5,7,9,11,13,15,14,12,10,8,6,4,2,
%U A382306 1,3,5,7,9,11,13,15,17,16,14,12,10,8,6,4,2,1
%N A382306 a(n) is the number of values m that satisfy floor(sqrt(m))=n and A382286(m)=1.
%C A382306 The sequence is quasiperiodic with quasiperiod 2*n+1.
%C A382306 The partial sum of the sequence up to N behaves like 2*N^(3/2)/3 for N large enough.
%F A382306 Let C(m) = floor(sqrt(A033677(m))) - floor(sqrt(A033676(m))), then
%F A382306 a(n) = |{m: n^2 <= m <= n^2+2*n and C(m)=0}|.
%F A382306 a(n^2)=3.
%F A382306 a(1)=3; a(2)=2; a(3)=1.
%F A382306 a(n^2+m) = 3+2*m for m=0..n-1.
%F A382306 a(n^2+n+m) = 2*(n-m) for m=0..n-1.
%F A382306 a(n^2+2*n) = 1.
%e A382306 a(1)=3 since C(1)=C(2)=C(3)=0.
%e A382306 a(2)=2 because C(4)..C(8) = 0,1,1,0,1 and only two arguments satisfy C(m)=0.
%e A382306 a(3)=1 because C(9)..C(15) = 0,1,2,1,2,1,1 and only one argument satisfies C(m)=0.
%e A382306 a(4)=3 since C(16)..C(24) = 0,3,1,3,0,1,2,3,0 and only three arguments satisfying C(m)=0.
%o A382306 (PARI) d(n) = if(n<2, 1, my(d=divisors(n)); d[(length(d)+1)\2]); \\ A033676
%o A382306 f(n) = my(k=1); while (sqrtint(n*k/d(n*k)) != sqrtint(d(n*k)), k++); k; \\ A382286
%o A382306 a(n) = #select(x->f(x)==1, [n^2..n^2+2*n]); \\ _Michel Marcus_, Mar 21 2025
%Y A382306 Cf A033676, A033677, A382286.
%K A382306 nonn
%O A382306 1,1
%A A382306 _Hassan Baloui_, Mar 21 2025