A359938 Number of divisors d of n such that d-1 is square.

1, 2, 1, 2, 2, 2, 1, 2, 1, 4, 1, 2, 1, 2, 2, 2, 2, 2, 1, 4, 1, 2, 1, 2, 2, 3, 1, 2, 1, 4, 1, 2, 1, 3, 2, 2, 2, 2, 1, 4, 1, 2, 1, 2, 2, 2, 1, 2, 1, 5, 2, 3, 1, 2, 2, 2, 1, 2, 1, 4, 1, 2, 1, 2, 3, 2, 1, 3, 1, 4, 1, 2, 1, 3, 2, 2, 1, 3, 1, 4, 1, 3, 1, 2, 3, 2, 1, 2, 1, 4

Offset: 1

Author

Seiichi Manyama, Jan 19 2023

Keywords

• nonn

Comments

The Cartesian equation for the Witch of Agnesi is given as y = 8*k^3/(x^2+4*k^2). If we set x = n, then a(n)-1 is the number of positive integer solutions in k such that y is a positive integer. Let d = m^2+1 be a divisor of n, then k = m*n/2 is a solution. - Thomas Scheuerle, Aug 07 2024

Crossrefs

Cf. A002522, A046951, A113319, A359937, A359944.

Programs

Mathematica

nmax = 100; Rest[CoefficientList[Series[Sum[x^(k^2 + 1)/(1 - x^(k^2 + 1)), {k, 0, Sqrt[nmax]}], {x, 0, nmax}], x]] (* Vaclav Kotesovec, Jan 19 2023 *)
Table[Sum[Floor[n/(k^2 + 1)] - Floor[(n-1)/(k^2 + 1)], {k, 0, Sqrt[n]}], {n, 1, 100}] (* Vaclav Kotesovec, Jan 19 2023 *)

PARI

a(n) = sumdiv(n, d, issquare(d-1));

PARI

my(N=100, x='x+O('x^N)); Vec(sum(k=0, sqrtint(N), x^(k^2+1)/(1-x^(k^2+1))))

Formula

G.f.: Sum_{k>=0} x^(k^2+1)/(1 - x^(k^2+1)).

From Vaclav Kotesovec, Jan 19 2023: (Start)

a(n) = Sum_{k=0..n} (floor(n/(k^2 + 1)) - floor((n-1)/(k^2 + 1))).

Sum_{k=1..n} a(k) = Sum_{k=0..n} floor(n/(k^2 + 1)).

Sum_{k=1..n} a(k) ~ c*n, where c = A113319 = (1 + Pi*coth(Pi))/2 = 2.0766... (End)