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 1-4 of 4 results.

A330570 Partial sums of A097988 (d_3(n)^2).

Original entry on oeis.org

1, 10, 19, 55, 64, 145, 154, 254, 290, 371, 380, 704, 713, 794, 875, 1100, 1109, 1433, 1442, 1766, 1847, 1928, 1937, 2837, 2873, 2954, 3054, 3378, 3387, 4116, 4125, 4566, 4647, 4728, 4809, 6105, 6114, 6195, 6276, 7176, 7185, 7914, 7923, 8247, 8571, 8652, 8661, 10686, 10722, 11046, 11127, 11451, 11460, 12360
Offset: 1

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Author

N. J. A. Sloane, Jan 08 2020

Keywords

Comments

This and the following sequences (and continuing in A331071) were inspired by the papers of Hooley, Indlekofer, Motohashi, Redmond, Titchmarsh, etc.

Crossrefs

Programs

  • Mathematica
    Accumulate[a[n_]:=DivisorSum[n, DivisorSigma[0, #]&]^2; Array[a, 60]] (* Vincenzo Librandi, Jan 11 2020 *)
  • PARI
    lista(nmax) = {my(s = 0); for(n = 1, nmax, s += vecprod(apply(e -> (e+1)*(e+2)/2, factor(n)[,2]))^2; print1(s, ", "));} \\ Amiram Eldar, Apr 19 2024

Formula

a(n) ~ c * n * log(n)^8 /8!, where c = Product_{p prime} ((1-1/p)^4 * (1 + 4/p + 1/p^2)) = 0.049321673579400091761... (Titchmarsh, 1942). - Amiram Eldar, Apr 19 2024

A344080 a(n) = Sum_{d|n} tau(d)^n, where tau(n) is the number of divisors of n.

Original entry on oeis.org

1, 5, 9, 98, 33, 4225, 129, 72354, 20196, 1050625, 2049, 2194099186, 8193, 268468225, 1073807361, 156925970179, 131073, 101629064089930, 524289, 3657261440572306, 4398050705409, 17592194433025, 8388609, 4727105427440383342818, 847322163876, 4503599761588225
Offset: 1

Views

Author

Seiichi Manyama, May 09 2021

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := DivisorSum[n, DivisorSigma[0, #]^n &]; Array[a, 26] (* Amiram Eldar, May 09 2021 *)
  • PARI
    a(n) = sumdiv(n, d, numdiv(d)^n);
    
  • PARI
    my(N=40, x='x+O('x^N)); Vec(sum(k=1, N, (numdiv(k)*x)^k/(1-(numdiv(k)*x)^k)))

Formula

G.f.: Sum_{k >= 1} (tau(k) * x)^k/(1 - (tau(k) * x)^k).
If p is prime, a(p) = 1 + 2^p.

A344081 a(n) = Sum_{d|n} tau(d)^d, where tau(n) is the number of divisors of n.

Original entry on oeis.org

1, 5, 9, 86, 33, 4109, 129, 65622, 19692, 1048613, 2049, 2176786526, 8193, 268435589, 1073741865, 152587956247, 131073, 101559956692208, 524289, 3656158441111670, 4398046511241, 17592186046469, 8388609, 4722366482871822065758
Offset: 1

Views

Author

Seiichi Manyama, May 09 2021

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := DivisorSum[n, DivisorSigma[0, #]^# &]; Array[a, 24] (* Amiram Eldar, May 09 2021 *)
  • PARI
    a(n) = sumdiv(n, d, numdiv(d)^d);
    
  • PARI
    my(N=40, x='x+O('x^N)); Vec(sum(k=1, N, (numdiv(k)*x)^k/(1-x^k)))

Formula

G.f.: Sum_{k >= 1} (tau(k) * x)^k/(1 - x^k).
If p is prime, a(p) = 1 + 2^p.

A344082 a(n) = n * Sum_{d|n} tau(d)^3 / d, where tau(n) is the number of divisors of n.

Original entry on oeis.org

1, 10, 11, 47, 13, 110, 15, 158, 60, 130, 19, 517, 21, 150, 143, 441, 25, 600, 27, 611, 165, 190, 31, 1738, 92, 210, 244, 705, 37, 1430, 39, 1098, 209, 250, 195, 2820, 45, 270, 231, 2054, 49, 1650, 51, 893, 780, 310, 55, 4851, 132, 920, 275, 987, 61, 2440, 247, 2370, 297, 370, 67, 6721, 69
Offset: 1

Views

Author

Seiichi Manyama, May 09 2021

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := n * DivisorSum[n, DivisorSigma[0, #]^3/# &]; Array[a, 61] (* Amiram Eldar, May 09 2021 *)
  • PARI
    a(n) = n*sumdiv(n, d, numdiv(d)^3/d);
    
  • PARI
    my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, numdiv(k)^3*x^k/(1-x^k)^2))

Formula

G.f.: Sum_{k >= 1} tau(k)^3 * x^k/(1 - x^k)^2.
If p is prime, a(p) = 8 + p.
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = zeta(2)^4 * Product_{p prime} (1 + 4/p^2 + 1/p^4) = 31.237542262502... . - Amiram Eldar, Dec 22 2023
From Peter Bala, Jan 25 2024: (Start)
a(n) = Sum_{d|n, e|n} gcd(d, e) * tau(n/d) * tau(n/e) (the sum is a multiplicative function of n - see Tóth).
Multiplicative: a(p^k) = ( p^(k+2)*(p^2 + 4*p + 1) - p^3*(k + 2)^3 + p^2*(3*k^3 + 15*k^2 + 21*k + 5) - p*(3*k^3 + 12*k^2 + 12*k + 4) + (k + 1)^3 ) / (p - 1)^4. (End)
Showing 1-4 of 4 results.