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.

A064945 a(n) = Sum_{i|n, j|n, j >= i} i.

Original entry on oeis.org

1, 4, 5, 11, 7, 22, 9, 26, 18, 30, 13, 64, 15, 38, 38, 57, 19, 82, 21, 87, 48, 54, 25, 156, 38, 62, 58, 109, 31, 179, 33, 120, 68, 78, 68, 244, 39, 86, 78, 213, 43, 224, 45, 153, 143, 102, 49, 348, 66, 166, 98, 175, 55, 268, 96, 267, 108, 126, 61, 542, 63, 134, 181
Offset: 1

Views

Author

Vladeta Jovovic, Oct 28 2001

Keywords

Examples

			a(6) = dot_product(4,3,2,1)*(1,2,3,6) = 4*1+3*2+2*3+1*6 = 22.
		

Crossrefs

Programs

  • Haskell
    a064945 = sum . zipWith (*) [1..] . reverse . a027750_row'
    -- Reinhard Zumkeller, Jul 14 2015
    
  • Maple
    with(numtheory): seq(add((tau(n)-i+1)*sort(convert(divisors(n),'list'))[i],i=1..tau(n)), n=1..200);
  • Mathematica
    A064945[n_] := #.Range[Length[#], 1, -1] & [Divisors[n]];
    Array[A064945, 100] (* Paolo Xausa, Aug 07 2025 *)
  • PARI
    a(n) = my(d=divisors(n), t=length(d)); sum(i=1, t, (t - i + 1)*d[i]); \\ Harry J. Smith, Oct 01 2009
    
  • PARI
    a(n)=my(d=divisors(n)); sum(i=1,#d,(#d+1-i)*d[i]) \\ Charles R Greathouse IV, Jun 10 2015
    
  • Python
    from sympy import divisors, divisor_sigma
    def A064945(n): return (divisor_sigma(n,0)+1)*divisor_sigma(n)-sum(a*b for a, b in enumerate(divisors(n),1)) # Chai Wah Wu, Aug 07 2025

Formula

a(n) = Sum_{i=1..tau(n)} (tau(n)-i+1)*d_i, where {d_i}, i=1..tau(n), is the increasing sequence of divisors of n.
a(n) = Sum_{i=1..A000005(n)} (A000005(n)-i+1)*A027750(n, i). - Michel Marcus, Jun 10 2015
From Ridouane Oudra, Aug 02 2025: (Start)
a(n) = Sum_{d|n} d*A135539(n,d).
a(n) = A064947(n) + A000203(n).
a(n) = (A064949(n) + A000203(n))/2.
a(n) = A064949(n) - A064947(n).
a(n) = A337360(n) - A064944(n).
a(n) = A064840(n) - A064946(n). (End)

A064944 a(n) = Sum_{i|n, j|n, j >= i} j.

Original entry on oeis.org

1, 5, 7, 17, 11, 38, 15, 49, 34, 60, 23, 132, 27, 82, 82, 129, 35, 191, 39, 207, 112, 126, 47, 384, 86, 148, 142, 283, 59, 469, 63, 321, 172, 192, 172, 666, 75, 214, 202, 597, 83, 640, 87, 435, 403, 258, 95, 1016, 162, 485, 262, 511, 107, 812, 264, 813, 292, 324
Offset: 1

Views

Author

Vladeta Jovovic, Oct 28 2001

Keywords

Examples

			a(6) = max(1,1)+max(1,2)+max(1,3)+max(1,6)+max(2,2)+max(2,3)+max(2,6)+max(3,3)+max(3,6)+max(6,6)=38, or a(6) = dot_product(1,2,3,4)*(1,2,3,6)=1*1+2*2+3*3+4*6=38.
		

Crossrefs

Programs

  • Haskell
    a064944 = sum . zipWith (*) [1..] . a027750_row'
    -- Reinhard Zumkeller, Jul 14 2015
    
  • Maple
    with(numtheory): seq(add(i*sort(convert(divisors(n),'list'))[i],i=1..tau(n)), n=1..200);
  • Mathematica
    A064944[n_] := #.Range[Length[#]] & [Divisors[n]];
    Array[A064944, 100] (* Paolo Xausa, Aug 07 2025 *)
  • PARI
    a(n) = my(d=divisors(n)); sum(i=1, length(d), i*d[i]); \\ Harry J. Smith, Sep 30 2009
    
  • Python
    from sympy import divisors
    def A064944(n): return sum(a*b for a, b in enumerate(divisors(n),1)) # Chai Wah Wu, Aug 07 2025

Formula

a(n) = Sum_{i=1..tau(n)} i*d_i, where {d_i}, i=1..tau(n) is the increasing sequence of divisors of n.
a(n) = Sum_{i=1..A000005(n)} i*A027750(n, i). - Michel Marcus, Jun 10 2015
From Ridouane Oudra, Aug 01 2025: (Start)
a(n) = Sum_{d|n} (n/d)*A135539(n,d).
a(n) = A064946(n) + A000203(n).
a(n) = (A064948(n) + A000203(n))/2.
a(n) = A337360(n) - A064945(n).
a(n) = A064948(n) - A064946(n).
a(n) = A064840(n) - A064947(n). (End)

A337297 a(n) = sigma(n)*(tau(n) - 1).

Original entry on oeis.org

0, 3, 4, 14, 6, 36, 8, 45, 26, 54, 12, 140, 14, 72, 72, 124, 18, 195, 20, 210, 96, 108, 24, 420, 62, 126, 120, 280, 30, 504, 32, 315, 144, 162, 144, 728, 38, 180, 168, 630, 42, 672, 44, 420, 390, 216, 48, 1116, 114, 465, 216, 490, 54, 840, 216, 840, 240, 270, 60, 1848, 62, 288
Offset: 1

Views

Author

Wesley Ivan Hurt, Aug 21 2020

Keywords

Comments

Original name was: Sum of the coordinates of all pairs of divisors of n, (d1,d2), such that d1 < d2.
If n = p where p is prime, the only pair of divisors of p such that d1 < d2 is (1,p), whose coordinate sum is a(p) = p + 1. - Wesley Ivan Hurt, May 21 2021

Examples

			a(3) = 4; The divisors of 3 are {1,3}. If we form all ordered pairs (d1,d2) such that d1 < d2, we have: (1,3). The sum of the coordinates gives 1+3 = 4.
a(4) = 14; The divisors of 4 are {1,2,4}. If we form all ordered pairs (d1,d2) such that d1<d2, we have: (1,2), (1,4), (2,4). The sum of all the coordinates gives 1+2+1+4+2+4 = 14.
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[Sum[(i + k)*(1 - Ceiling[n/k] + Floor[n/k]) (1 - Ceiling[n/i] + Floor[n/i]), {i, k - 1}], {k, n}], {n, 100}]
  • PARI
    a(n) = my(d = divisors(n)); sum(i=1, #d, sum(j=1, i-1, d[i]+d[j])); \\ Michel Marcus, Aug 22 2020

Formula

a(n) = Sum_{d1|n, d2|n, d1
a(p^k) = k*(p^(k+1)-1)/(p-1) for p prime and k >= 1. - Wesley Ivan Hurt, Aug 23 2025

Extensions

New name using formula from Ridouane Oudra, Jul 31 2025

A334954 a(n) is 1 plus the number of divisors of n.

Original entry on oeis.org

2, 3, 3, 4, 3, 5, 3, 5, 4, 5, 3, 7, 3, 5, 5, 6, 3, 7, 3, 7, 5, 5, 3, 9, 4, 5, 5, 7, 3, 9, 3, 7, 5, 5, 5, 10, 3, 5, 5, 9, 3, 9, 3, 7, 7, 5, 3, 11, 4, 7, 5, 7, 3, 9, 5, 9, 5, 5, 3, 13, 3, 5, 7, 8, 5, 9, 3, 7, 5, 9, 3, 13, 3, 5, 7, 7, 5, 9, 3, 11, 6, 5, 3, 13, 5, 5, 5, 9, 3, 13, 5, 7, 5, 5, 5, 13, 3, 7, 7, 10, 3, 9, 3, 9
Offset: 1

Author

David A. Corneth and Omar E. Pol, Aug 25 2020

Keywords

Comments

a(n) is the number of times that every divisor of n occurs in the coordinates of divisors of n mentioned in A337360 (Corneth).
a(n) = 3 if and only if n is prime.
a(n) is even if and only if n is a square.
a(n) is the number of characteristic subgroups of the dihedral group D_2n. - Firdous Ahmad Mala, Dec 25 2021

Crossrefs

Partial sums give A156745.

Programs

  • Mathematica
    1 + DivisorSigma[0, Range[105]] (* Michael De Vlieger, Sep 11 2020 *)
  • PARI
    a(n) = numdiv(n) + 1

Formula

a(n) = 1 + A000005(n).
a(n) = A337360(n)/A000203(n).
a(n) = A212356(n) for n >= 3. - Ilya Gutkovskiy, Aug 27 2020
Showing 1-4 of 4 results.