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.

A350125 a(n) = Sum_{k=1..n} k^2 * floor(n/k)^n.

Original entry on oeis.org

1, 8, 40, 345, 3303, 50225, 833569, 17045934, 388654659, 10039636255, 285508661853, 8924967326015, 302927979357701, 11114722212099135, 437913155876193839, 18447871416712820782, 827249276230172525622, 39347009369000530723017
Offset: 1

Views

Author

Seiichi Manyama, Dec 15 2021

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := Sum[k^2 * Floor[n/k]^n, {k, 1, n}]; Array[a, 18] (* Amiram Eldar, Oct 04 2023 *)
  • PARI
    a(n) = sum(k=1, n, k^2*(n\k)^n);
    
  • PARI
    a(n) = sum(k=1, n, k^2*sumdiv(k, d, (d^n-(d-1)^n)/d^2));

Formula

a(n) = Sum_{k=1..n} k^2 * Sum_{d|k} (d^n - (d - 1)^n)/d^2.
a(n) = [x^n] (1/(1 - x)) * Sum_{k>=1} (k^n - (k - 1)^n) * x^k * (1 + x^k)/(1 - x^k)^3.
a(n) ~ n^n. - Vaclav Kotesovec, Dec 16 2021

A350106 Square array T(n,k), n >= 1, k >= 1, read by antidiagonals downwards, where T(n,k) = Sum_{j=1..n} j * floor(n/j)^k.

Original entry on oeis.org

1, 1, 4, 1, 6, 8, 1, 10, 14, 15, 1, 18, 32, 31, 21, 1, 34, 86, 87, 45, 33, 1, 66, 248, 295, 153, 81, 41, 1, 130, 734, 1095, 669, 309, 101, 56, 1, 258, 2192, 4231, 3201, 1521, 443, 150, 69, 1, 514, 6566, 16647, 15765, 8373, 2633, 722, 191, 87, 1, 1026, 19688, 66055, 78393, 48321, 17411, 4746, 1005, 253, 99
Offset: 1

Views

Author

Seiichi Manyama, Dec 14 2021

Keywords

Examples

			Square array begins:
   1,   1,   1,    1,     1,      1,      1, ...
   4,   6,  10,   18,    34,     66,    130, ...
   8,  14,  32,   86,   248,    734,   2192, ...
  15,  31,  87,  295,  1095,   4231,  16647, ...
  21,  45, 153,  669,  3201,  15765,  78393, ...
  33,  81, 309, 1521,  8373,  48321, 284709, ...
  41, 101, 443, 2633, 17411, 119321, 828323, ...
		

Crossrefs

Columns k=1..3 give A024916, A350107, A350108.
T(n,n) gives A350109.

Programs

  • Mathematica
    T[n_, k_] := Sum[j * Floor[n/j]^k, {j, 1, n}]; Table[T[k, n - k + 1], {n, 1, 11}, {k, 1, n}] // Flatten (* Amiram Eldar, Dec 14 2021 *)
  • PARI
    T(n, k) = sum(j=1, n, j*(n\j)^k);
    
  • PARI
    T(n, k) = sum(j=1, n, j*sumdiv(j, d, (d^k-(d-1)^k)/d));

Formula

G.f. of column k: (1/(1 - x)) * Sum_{j>=1} (j^k - (j - 1)^k) * x^j/(1 - x^j)^2.
T(n,k) = Sum_{j=1..n} j * Sum_{d|j} (d^k - (d - 1)^k)/d.

A356131 a(n) = Sum_{k=1..n} (k - 1)^n * binomial(floor(n/k)+1,2).

Original entry on oeis.org

0, 1, 9, 100, 1302, 20648, 377022, 7921039, 186926431, 4916562309, 142373072781, 4506381442625, 154721361953489, 5729251983077521, 227585590018322461, 9654855432715969784, 435659531345223039702, 20836069677785611552293
Offset: 1

Views

Author

Seiichi Manyama, Jul 27 2022

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := Sum[(k - 1)^n * Binomial[Floor[n/k]+1, 2], {k, 1, n}]; Array[a, 18] (* Amiram Eldar, Jul 28 2022 *)
  • PARI
    a(n) = sum(k=1, n, (k-1)^n*binomial((n\k)+1, 2));
    
  • PARI
    a(n) = sum(k=1, n, k*(sigma(k, n-1)-(n\k)^n));
    
  • PARI
    a(n) = sum(k=1, n, k*sumdiv(k, d, (d-1)^n/d));

Formula

a(n) = Sum_{k=1..n} k * (sigma_{n-1}(k) - floor(n/k)^n) = A356129(n) - A350109(n).
a(n) = Sum_{k=1..n} k * Sum_{d|k} (d - 1)^n / d.
a(n) = [x^n] (1/(1-x)) * Sum_{k>=1} (k - 1)^n * x^k/(1 - x^k)^2.

A356238 a(n) = Sum_{k=1..n} (k * floor(n/k))^n.

Original entry on oeis.org

1, 8, 62, 849, 8541, 206345, 2581403, 76623522, 1617299079, 49463993875, 952905453423, 59000021366675, 1198427462876421, 54128102218676115, 2321105129608323165, 117387839988330848902, 3205342976298888473968, 268263812478494295219717
Offset: 1

Views

Author

Seiichi Manyama, Jul 30 2022

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := Sum[(k * Floor[n/k])^n, {k, 1, n}]; Array[a, 18] (* Amiram Eldar, Jul 30 2022 *)
  • PARI
    a(n) = sum(k=1, n, (k*(n\k))^n);
    
  • PARI
    a(n) = sum(k=1, n, k^n*sumdiv(k, d, 1-(1-1/d)^n));

Formula

a(n) = Sum_{k=1..n} k^n * Sum_{d|k} (1 - (1 - 1/d)^n).
Showing 1-4 of 4 results.