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

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, 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, 1, 3, 5, 7, 9, 11, 13, 15, 17, 16, 14, 12, 10, 8, 6, 4, 2, 1

Offset: 1

Author

Hassan Baloui, Mar 21 2025

Keywords

• nonn

Comments

The sequence is quasiperiodic with quasiperiod 2*n+1.

The partial sum of the sequence up to N behaves like 2*N^(3/2)/3 for N large enough.

Examples

			a(1)=3 since C(1)=C(2)=C(3)=0.
a(2)=2 because C(4)..C(8) = 0,1,1,0,1 and only two arguments satisfy C(m)=0.
a(3)=1 because C(9)..C(15) = 0,1,2,1,2,1,1 and only one argument satisfies C(m)=0.
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.
		

Crossrefs

Cf A033676, A033677, A382286.

Programs

PARI

d(n) = if(n<2, 1, my(d=divisors(n)); d[(length(d)+1)\2]); \\ A033676
f(n) = my(k=1); while (sqrtint(n*k/d(n*k)) != sqrtint(d(n*k)), k++); k; \\ A382286
a(n) = #select(x->f(x)==1, [n^2..n^2+2*n]); \\ Michel Marcus, Mar 21 2025

Formula

Let C(m) = floor(sqrt(A033677(m))) - floor(sqrt(A033676(m))), then

a(n) = |{m: n^2 <= m <= n^2+2*n and C(m)=0}|.

a(n^2)=3.

a(1)=3; a(2)=2; a(3)=1.

a(n^2+m) = 3+2*m for m=0..n-1.

a(n^2+n+m) = 2*(n-m) for m=0..n-1.

a(n^2+2*n) = 1.