cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A359943 a(n) = Sum_{d|n, d-1 is cube} d.

Original entry on oeis.org

1, 3, 1, 3, 1, 3, 1, 3, 10, 3, 1, 3, 1, 3, 1, 3, 1, 12, 1, 3, 1, 3, 1, 3, 1, 3, 10, 31, 1, 3, 1, 3, 1, 3, 1, 12, 1, 3, 1, 3, 1, 3, 1, 3, 10, 3, 1, 3, 1, 3, 1, 3, 1, 12, 1, 31, 1, 3, 1, 3, 1, 3, 10, 3, 66, 3, 1, 3, 1, 3, 1, 12, 1, 3, 1, 3, 1, 3, 1, 3, 10, 3, 1, 31, 1, 3
Offset: 1

Views

Author

Seiichi Manyama, Jan 19 2023

Keywords

Crossrefs

Cf. A001093, A113061, A359937, A359942, A359944.

Programs

  • Mathematica
    a[n_] := DivisorSum[n, # &, IntegerQ[Surd[#-1, 3]] &]; Array[a, 100] (* Amiram Eldar, Aug 09 2023 *)
  • PARI
    a(n) = sumdiv(n, d, ispower(d-1, 3)*d);

Formula

G.f.: Sum_{k>=0} (k^3+1) * x^(k^3+1)/(1 - x^(k^3+1)).

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

Original entry on oeis.org

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

Views

Author

Seiichi Manyama, Jan 19 2023

Keywords

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)
Showing 1-2 of 2 results.

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