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.

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A356243 a(n) = Sum_{k=1..n} k^2 * sigma_{n-2}(k).

Original entry on oeis.org

1, 9, 49, 447, 4607, 71009, 1210855, 24980627, 575624572, 14958422046, 427890493960, 13431874937840, 457651929853662, 16844143705998554, 665499756005678382, 28102799297908820326, 1262909308355648335240, 60183118566605371095996
Offset: 1

Views

Author

Seiichi Manyama, Jul 30 2022

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := Sum[k^2 * DivisorSigma[n - 2, k], {k, 1, n}]; Array[a, 18] (* Amiram Eldar, Jul 30 2022 *)
  • PARI
    a(n) = sum(k=1, n, k^2*sigma(k, n-2));
    
  • PARI
    a(n) = sum(k=1, n, k^n*sum(j=1, n\k, j^2));
    
  • Python
    from math import isqrt
    from sympy import bernoulli
    def A356243(n): return (((s:=isqrt(n))*(s+1)*(2*s+1))*((b:=bernoulli(n+1))-bernoulli(n+1, s+1)) + sum(k**n*(n+1)*((q:=n//k)*(q+1)*(2*q+1))+6*k**2*(bernoulli(n+1,q+1)-b) for k in range(1,s+1)))//(n+1)//6 # Chai Wah Wu, Oct 21 2023

Formula

a(n) = Sum_{k=1..n} k^n * Sum_{j=1..floor(n/k)} j^2 = Sum_{k=1..n} k^n * A000330(floor(n/k)).
a(n) = [x^n] (1/(1-x)) * Sum_{k>=1} k^n * x^k * (1 + x^k)/(1 - x^k)^3.

A356100 a(n) = Sum_{k=1..n} (k - 1)^n * floor(n/k).

Original entry on oeis.org

0, 1, 9, 99, 1301, 20581, 376891, 7914216, 186905206, 4915451602, 142368695176, 4506118905870, 154720069309364, 5729167232515112, 227585086051159866, 9654819212943764500, 435659280972794395356, 20836049921760968809231, 1052864549462731148832219
Offset: 1

Views

Author

Seiichi Manyama, Jul 26 2022

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sum[(k-1)^n Floor[n/k],{k,n}],{n,20}] (* Harvey P. Dale, Dec 14 2024 *)
  • PARI
    a(n) = sum(k=1, n, (k-1)^n*(n\k));
    
  • PARI
    a(n) = sum(k=1, n, sigma(k, n)-(n\k)^n);
    
  • PARI
    a(n) = sum(k=1, n, sumdiv(k, d, (d-1)^n));
    
  • Python
    def A356100(n): return sum((k-1)**n*(n//k) for k in range(2,n+1)) # Chai Wah Wu, Jul 26 2022

Formula

a(n) = A319194(n) - A332469(n).
a(n) = Sum_{k=1..n} Sum_{d|k} (d - 1)^n.
a(n) = [x^n] (1/(1-x)) * Sum_{k>=1} (k - 1)^n * x^k/(1 - x^k).

A356130 a(n) = Sum_{k=1..n} sigma_{n-1}(k).

Original entry on oeis.org

1, 4, 16, 111, 999, 12513, 185683, 3316418, 67810767, 1576561677, 40862702931, 1171104916405, 36722498575799, 1251419967587955, 46034784688102781, 1818440444592581068, 76763036794222996512, 3448830049286378614987, 164309958491233496689189
Offset: 1

Views

Author

Seiichi Manyama, Jul 27 2022

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := Sum[DivisorSigma[n-1, k], {k, 1, n}]; Array[a, 19] (* Amiram Eldar, Jul 28 2022 *)
  • PARI
    a(n) = sum(k=1, n, sigma(k, n-1));
    
  • PARI
    a(n) = sum(k=1, n, k^(n-1)*(n\k));
    
  • Python
    from math import isqrt
    from sympy import bernoulli
    def A350130(n): return (((s:=isqrt(n))+1)*((b:=bernoulli(n))-bernoulli(n, s+1))+sum(k**(n-1)*n*((q:=n//k)+1)-b+bernoulli(n, q+1) for k in range(1,s+1)))//n if n>1 else 1 # Chai Wah Wu, Oct 21 2023

Formula

a(n) = Sum_{k=1..n} k^(n-1) * floor(n/k).
a(n) = [x^n] (1/(1-x)) * Sum_{k>=1} k^(n-1) * x^k/(1 - x^k).

A308652 a(n) = Product_{k=1..n} sigma(n,k).

Original entry on oeis.org

1, 1, 5, 252, 380562, 26605273464, 146392210728465000, 84641321148614770425516288, 7097143900835489590932722296959504144, 109983275218947201453245400551817117367706036248320, 397899007017966277799025689101644536884667639093655295898437500000
Offset: 0

Views

Author

Vaclav Kotesovec, Aug 20 2019

Keywords

Crossrefs

Programs

  • Maple
    with(NumberTheory): seq(product(sigma[n](k), k = 1..n), n = 0..10);
  • Mathematica
    Table[Product[DivisorSigma[n, k], {k, 1, n}], {n, 1, 10}]
  • PARI
    a(n) = prod(k=1, n, sigma(k, n));
    for(n=1, 10, print1(a(n), ", "))

Formula

a(n) ~ (n!)^n.
a(n) ~ 2^(n/2) * Pi^(n/2) * n^(n*(2*n+1)/2) / exp(n^2-1/12).

Extensions

a(0)=1 prepended by Alois P. Heinz, Aug 23 2019

A319278 Square array sigma_k(n) read down antidiagonals: sum of the k-th powers of the divisors of n.

Original entry on oeis.org

1, 1, 3, 1, 5, 4, 1, 9, 10, 7, 1, 17, 28, 21, 6, 1, 33, 82, 73, 26, 12, 1, 65, 244, 273, 126, 50, 8, 1, 129, 730, 1057, 626, 252, 50, 15, 1, 257, 2188, 4161, 3126, 1394, 344, 85, 13, 1, 513, 6562, 16513, 15626, 8052, 2402, 585, 91, 18, 1, 1025, 19684, 65793, 78126, 47450, 16808, 4369, 757, 130, 12
Offset: 1

Views

Author

R. J. Mathar, Sep 16 2018

Keywords

Comments

Equals the square array A082771 without its first column.

Examples

			The array starts in row n=1 with columns k>=1 as:
     1      1      1      1      1      1       1        1
     3      5      9     17     33     65     129      257
     4     10     28     82    244    730    2188     6562
     7     21     73    273   1057   4161   16513    65793
     6     26    126    626   3126  15626   78126   390626
    12     50    252   1394   8052  47450  282252  1686434
     8     50    344   2402  16808 117650  823544  5764802
    15     85    585   4369  33825 266305 2113665 16843009
		

Crossrefs

Cf. A082771, A023887 (diagonal), A109974, A319194 (partial column sums).

Programs

  • Mathematica
    T[n_, k_] := DivisorSigma[k, n];
    Table[T[n-k+1, k], {n, 1, 11}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Dec 16 2021 *)

Formula

sigma_k(n) = sum_{d|n} d^k.

A332617 a(n) = Sum_{k=1..n} J_n(k), where J is the Jordan function, J_n(k) = k^n * Product_{p|k, p prime} (1 - 1/p^n).

Original entry on oeis.org

1, 4, 34, 336, 4390, 66312, 1197858, 24612000, 574002448, 14903406552, 427622607366, 13419501812640, 457579466056498, 16840326075104280, 665473192580864556, 28101209228393371200, 1262896789586657015796, 60182268296582518426368, 3031282541337682050032664
Offset: 1

Views

Author

Ilya Gutkovskiy, Feb 17 2020

Keywords

Crossrefs

Programs

  • Magma
    [&+[&+[MoebiusMu(k div d)*d^n:d in Divisors(k)]:k in [1..n]]:n in [1..20]]; // Marius A. Burtea, Feb 17 2020
  • Mathematica
    Table[Sum[Sum[MoebiusMu[k/d] d^n, {d, Divisors[k]}], {k, 1, n}], {n, 1, 19}]
    Table[SeriesCoefficient[(1/(1 - x)) Sum[Sum[MoebiusMu[k] j^n x^(k j), {j, 1, n}], {k, 1, n}], {x, 0, n}], {n, 1, 19}]

Formula

a(n) = [x^n] (1/(1 - x)) * Sum_{k>=1} Sum_{j>=1} mu(k) * j^n * x^(k*j).
Previous Showing 11-16 of 16 results.