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

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, 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, 13, 15, 18, 25, 24, 28, 22, 21, 18, 13, 11, 7, 3, 0, 3, 6, 11, 16, 20, 23, 27, 29, 30

Offset

1,2

Comments

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

Conjecture: Sum_{n=1..N} a(n) ~ 4*N^(3/2)/3 for N large enough.

Conjecture: Sum_{n=1..N} a(n) ~ 2*Sum_{n=1..N} A382306(n) for N large enough.

Links

Table of n, a(n) for n=1..80.

Formula

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))).

Example

a(1) = 0 because C(1) = 5, C(2) = 4, C(3) = 4.
a(2) = 3 because C(4)..C(8) = 2,1,2,1,1 and only three arguments satisfy C(m) = 1.
a(3) = 4 because C(9)..C(15) = 2,1,4,1,4,1,1 and only four arguments satisfies C(m) = 1.
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.

Prog

(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)) != 1, k++); k; \\ A381054
a(n) = #select(x->f(x)==1, [n^2..n^2+2*n]);

Crossrefs

Cf. A033676, A033677, A382286, A381054, A382306.

Sequence in context: A077451 A352770 A019829 * A384782 A200125 A091528

Adjacent sequences: A383112 A383113 A383114 * A383116 A383117 A383118

Keyword

nonn

Author

Hassan Baloui, Apr 16 2025

Status

approved