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.

Previous Showing 51-60 of 60 results.

A360159 a(n) is the sum of divisors of n that are odd squares.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 10, 1, 1, 1, 1, 1, 1, 1, 1, 10, 1, 1, 1, 1, 1, 1, 26, 1, 10, 1, 1, 1, 1, 1, 1, 1, 1, 10, 1, 1, 1, 1, 1, 1, 1, 1, 10, 1, 1, 1, 50, 26, 1, 1, 1, 10, 1, 1, 1, 1, 1, 1, 1, 1, 10, 1, 1, 1, 1, 1, 1, 1, 1, 10, 1, 1, 26, 1, 1, 1, 1, 1, 91, 1, 1
Offset: 1

Views

Author

Amiram Eldar, Jan 29 2023

Keywords

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := (p^(2*(1 + Floor[e/2])) - 1)/(p^2 - 1); f[2, e_] := 1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i, 1] == 2, 1, (f[i, 1]^(2*(f[i, 2]\2)+2)-1)/(f[i, 1]^2-1))); }

Formula

a(n) = Sum_{d|n, d odd square} d.
a(n) = (A035316(n) + A344300(n))/2.
Multiplicative with a(2^e) = 1, and for p > 2, a(p^e) = (p^(e+2)-1)/(p^2-1) for even e and a(p^e) = (p^(e+1)-1)/(p^2-1) for odd e.
Dirichlet g.f.: zeta(s)*zeta(2s-2)*(1-4^(1-s)).
Sum_{k=1..n} a(k) ~ c * n^(3/2), where c = zeta(3/2)/6 = 0.4353958914... .

A055928 Sum of square divisors of n! = sum of squares of divisors of the square root of largest square dividing n!.

Original entry on oeis.org

1, 1, 1, 5, 5, 210, 210, 850, 7735, 806806, 806806, 3229590, 3229590, 161479500, 1455090000, 23286770000, 23286770000, 838446909300, 838446909300, 83973923013750, 83973923013750, 10244818607677500, 10244818607677500
Offset: 1

Views

Author

Labos Elemer, Jul 21 2000

Keywords

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[OddQ[e], (p^(e+1)-1)/(p^2-1) , (p^(e+2)-1)/(p^2-1)]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n!]); Array[a, 23] (* Amiram Eldar, Aug 01 2019 *)

Formula

a(n) = A035316(n!) = A001157(A055772(n)) = A001157(A000188(n!)).

A065745 Sum of squares and twice squares dividing n.

Original entry on oeis.org

1, 3, 1, 7, 1, 3, 1, 15, 10, 3, 1, 7, 1, 3, 1, 31, 1, 30, 1, 7, 1, 3, 1, 15, 26, 3, 10, 7, 1, 3, 1, 63, 1, 3, 1, 70, 1, 3, 1, 15, 1, 3, 1, 7, 10, 3, 1, 31, 50, 78, 1, 7, 1, 30, 1, 15, 1, 3, 1, 7, 1, 3, 10, 127, 1, 3, 1, 7, 1, 3, 1, 150, 1, 3, 26, 7, 1, 3, 1, 31, 91, 3, 1, 7, 1, 3, 1, 15, 1, 30, 1
Offset: 1

Views

Author

Vladeta Jovovic, Dec 04 2001

Keywords

Crossrefs

Programs

  • Mathematica
    f[2, e_] := 2^(e+1) - 1; f[p_, e_] := If[OddQ[e], (p^(e+1)-1)/(p^2-1), (p^(e+2)-1)/(p^2-1)]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 13 2020 *)
  • PARI
    a(n) = sumdiv(n, d, issquare(d)*d + (1 - d%2)*issquare(d/2)*d) \\ Michel Marcus, Jun 17 2013

Formula

Multiplicative with a(2^e) = 2^(e+1)-1, a(p^e) = (p^(e+2)-1)/(p-1)/(p+1) for odd p and even e and a(p^e) = (p^(e+1)-1)/(p-1)/(p+1) for odd p and odd e.
From Amiram Eldar, Dec 15 2023: (Start)
Dirichlet g.f.: (1 + 1/2^(s-1)) * zeta(2*s-2) * zeta(s).
Sum_{k=1..n} a(k) ~ c * n^(3/2), where c = ((2+sqrt(2))/6) * zeta(3/2) = 1.4865345575818562471... . (End)

Extensions

More terms from Matthew Conroy, Jan 19 2002

A209311 Numbers whose sum of triangular divisors is also a divisor and greater than 1.

Original entry on oeis.org

285, 1302, 1425, 1820, 2508, 3640, 3720, 4845, 4956, 5016, 5415, 7125, 7280, 9100, 9114, 9912, 11685, 12255, 12740, 14508, 15105, 16815, 17385, 18200, 19095, 19824, 20235, 20805, 22134, 22515, 23655, 23660, 24021, 24738, 25365, 25480, 27075, 27588, 27645
Offset: 1

Views

Author

Antonio Roldán, Jan 18 2013

Keywords

Examples

			285 is in the sequence because its divisors being 1, 3, 5, 15, 19, 57, 95, 285, of which 1, 3 and 15 are triangular, these add up to 19.
1302 is in sequence because the sum of triangular divisors 21 + 6 + 3 + 1 = 31 is divisor of 1302.
		

Crossrefs

Programs

  • Mathematica
    TriangularQ[n_] := IntegerQ[Sqrt[1 + 8*n]]; fQ[n_] := Module[{tri = Total[Select[Divisors[n], TriangularQ]]}, tri > 1 && Mod[n, tri] == 0]; Select[Range[28000], fQ] (* T. D. Noe, Jan 23 2013 *)
  • PARI
    istriangular(n)=issquare(8*n+1)
    {t=0; for(n=1, 10^7, k=sumdiv(n, d, istriangular(d)*d); if(n/k==n\k&&k>>1, t+=1; write("b209311.txt",t," ",n)))}

A342228 Total sum of parts which are squares in all partitions of n.

Original entry on oeis.org

0, 1, 2, 4, 11, 16, 27, 42, 69, 108, 158, 229, 334, 469, 656, 903, 1255, 1685, 2283, 3032, 4033, 5290, 6936, 8986, 11650, 14969, 19172, 24402, 30998, 39110, 49260, 61712, 77155, 96000, 119209, 147394, 181958, 223713, 274533, 335792, 409980, 498981, 606273, 734572
Offset: 0

Views

Author

Ilya Gutkovskiy, Mar 06 2021

Keywords

Examples

			For n = 4 we have:
---------------------------------
Partitions        Sum of parts
.              which are squares
---------------------------------
4 ................... 4
3 + 1 ............... 1
2 + 2 ............... 0
2 + 1 + 1 ........... 2
1 + 1 + 1 + 1 ....... 4
---------------------------------
Total .............. 11
So a(4) = 11.
		

Crossrefs

Programs

  • Mathematica
    nmax = 43; CoefficientList[Series[Sum[k^2 x^(k^2)/(1 - x^(k^2)), {k, 1, Floor[nmax^(1/2)] + 1}]/Product[(1 - x^j), {j, 1, nmax}], {x, 0, nmax}], x]
    Table[Sum[DivisorSum[k, # &, IntegerQ[#^(1/2)] &] PartitionsP[n - k], {k, 1, n}], {n, 0, 43}]

Formula

G.f.: Sum_{k>=1} k^2*x^(k^2)/(1 - x^(k^2)) / Product_{j>=1} (1 - x^j).
a(n) = Sum_{k=1..n} A035316(k) * A000041(n-k).

A349330 a(n) = Sum_{d|n} d^c(d), where c is the characteristic function of squares (A010052).

Original entry on oeis.org

1, 2, 2, 6, 2, 4, 2, 7, 11, 4, 2, 9, 2, 4, 4, 23, 2, 14, 2, 9, 4, 4, 2, 11, 27, 4, 12, 9, 2, 8, 2, 24, 4, 4, 4, 55, 2, 4, 4, 11, 2, 8, 2, 9, 14, 4, 2, 28, 51, 30, 4, 9, 2, 16, 4, 11, 4, 4, 2, 15, 2, 4, 14, 88, 4, 8, 2, 9, 4, 8, 2, 58, 2, 4, 30, 9, 4, 8, 2, 28, 93, 4, 2, 15, 4, 4, 4
Offset: 1

Views

Author

Wesley Ivan Hurt, Nov 15 2021

Keywords

Comments

For each divisor d of n, add d if d is a square, otherwise add 1 [see example].
Inverse Möbius transform of n^c(n), where c = A010052. - Wesley Ivan Hurt, Mar 31 2025

Examples

			The divisors of 12 are 1, 2, 3, 4, 6, and 12 with squares 1 and 4, so a(12) = 1 + 1 + 1 + 4 + 1 + 1 = 9 (respectively).
		

Crossrefs

Programs

  • Mathematica
    a[n_] := DivisorSum[n, If[IntegerQ @ Sqrt[#], #, 1] &]; Array[a, 100] (* Amiram Eldar, Nov 15 2021 *)
  • PARI
    a(n) = sumdiv(n, d, if (issquare(d), d, 1)); \\ Michel Marcus, Nov 15 2021
    
  • PARI
    a(n) = {my(f = factor(n), cf = f, res); cf[,2]\=2; res = numdiv(f)-prod(i = 1, #f~, cf[i, 2]+1); res+=prod(i = 1, #f~, ((f[i,1]^(2*(cf[i,2]+1))-1)/(f[i,1]^2-1))); res } \\ David A. Corneth, Nov 16 2021

Formula

a(p) = 2 iff p is prime. - Wesley Ivan Hurt, Nov 28 2021
a(n) = A035316(n) + A056595(n). - R. J. Mathar, Aug 18 2024

A361793 Sum of the squares d^2 of the divisors d satisfying d^3|n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 10, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 10, 1, 5, 1, 1, 1, 1, 1, 1, 1, 21, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 5
Offset: 1

Views

Author

R. J. Mathar, Mar 24 2023

Keywords

Comments

The Mobius transform is 1, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, ... = n^(2/3)*A010057(n).

Crossrefs

Programs

  • Maple
    gsigma := proc(n,z,k)
        local a,d ;
        a := 0 ;
        for d in numtheory[divisors](n) do
            if modp(n,d^k) = 0 then
                a := a+d^z ;
            end if ;
        end do:
        a ;
    end proc:
    seq( gsigma(n,2,3),n=1..80) ;
  • Mathematica
    f[p_, e_] := (p^(2*(Floor[e/3] + 1)) - 1)/(p^2 - 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Mar 24 2023 *)
  • PARI
    a(n) = sumdiv(n, d, if (ispower(d, 3), sqrtnint(d, 3)^2)); \\ Michel Marcus, Mar 24 2023
    
  • PARI
    for(n=1, 100, print1(direuler(p=2, n, (1/((1-X)*(1-p^2*X^3))))[n], ", ")) \\ Vaclav Kotesovec, Jun 26 2024
    
  • Python
    from math import prod
    from sympy import factorint
    def A361793(n): return prod((p**(e//3+1<<1)-1)//(p**2-1) for p, e in factorint(n).items()) # Chai Wah Wu, Mar 24 2023

Formula

a(n) = Sum_{d^3|n} d^2.
Multiplicative with a(p^e) = (p^(2*(floor(e/3) + 1)) - 1)/(p^2 - 1). - Amiram Eldar, Mar 24 2023
G.f.: Sum_{k>=1} k^2 * x^(k^3) / (1 - x^(k^3)). - Ilya Gutkovskiy, Jun 05 2024
From Vaclav Kotesovec, Jun 26 2024: (Start)
Dirichlet g.f.: zeta(s)*zeta(3*s-2).
Sum_{k=1..n} a(k) ~ n*(log(n) + 4*gamma - 1)/3, where gamma is the Euler-Mascheroni constant A001620. (End)

A361794 Sum of the cubes d^3 of the divisors d satisfying d^2|n.

Original entry on oeis.org

1, 1, 1, 9, 1, 1, 1, 9, 28, 1, 1, 9, 1, 1, 1, 73, 1, 28, 1, 9, 1, 1, 1, 9, 126, 1, 28, 9, 1, 1, 1, 73, 1, 1, 1, 252, 1, 1, 1, 9, 1, 1, 1, 9, 28, 1, 1, 73, 344, 126, 1, 9, 1, 28, 1, 9, 1, 1, 1, 9, 1, 1, 28, 585, 1, 1, 1, 9, 1, 1, 1, 252, 1, 1, 126, 9, 1, 1, 1, 73
Offset: 1

Views

Author

R. J. Mathar, Mar 24 2023

Keywords

Comments

The Mobius transform is 1, 0, 0, 8, 0, 0, 0, 0, 27, 0, 0, ... = n^(3/2)*A010052(n).

Crossrefs

Programs

  • Maple
    gsigma := proc(n,z,k)
        local a,d ;
        a := 0 ;
        for d in numtheory[divisors](n) do
            if modp(n,d^k) = 0 then
                a := a+d^z ;
            end if ;
        end do:
        a ;
    end proc:
    seq( gsigma(n,3,2),n=1..80) ;
  • Mathematica
    f[p_, e_] := (p^(3*(Floor[e/2] + 1)) - 1)/(p^3 - 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Mar 24 2023 *)
  • PARI
    a(n) = sumdiv(n, d, if (issquare(d), sqrtint(d)^3)); \\ Michel Marcus, Mar 24 2023
    
  • Python
    from math import prod
    from sympy import factorint
    def A361794(n): return prod((p**(3*(e>>1)+3)-1)//(p**3-1) for p, e in factorint(n).items()) # Chai Wah Wu, Mar 24 2023

Formula

a(n) = Sum_{d^2|n} d^3.
Multiplicative with a(p^e) = (p^(3*(floor(e/2) + 1)) - 1)/(p^3 - 1). - Amiram Eldar, Mar 24 2023
G.f.: Sum_{k>=1} k^3 * x^(k^2) / (1 - x^(k^2)). - Ilya Gutkovskiy, Jun 05 2024

A367989 The sum of square divisors of the largest unitary divisor of n that is a square.

Original entry on oeis.org

1, 1, 1, 5, 1, 1, 1, 1, 10, 1, 1, 5, 1, 1, 1, 21, 1, 10, 1, 5, 1, 1, 1, 1, 26, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 50, 1, 1, 1, 1, 1, 1, 1, 5, 10, 1, 1, 21, 50, 26, 1, 5, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 10, 85, 1, 1, 1, 5, 1, 1, 1, 10, 1, 1, 26, 5, 1, 1, 1, 21, 91, 1
Offset: 1

Views

Author

Amiram Eldar, Dec 07 2023

Keywords

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[EvenQ[e], (p^(e + 2) - 1)/(p^2 - 1), 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,2]%2, 1, (f[i,1]^(f[i,2] + 2) - 1)/(f[i,1]^2 - 1)));}

Formula

a(n) = A035316(A350388(n)).
Multiplicative with a(p^e) = (p^(e+2)-1)/(p^2-1) if e is even and 1 otherwise.
a(n) >= 1, with equality if and only if n is an exponentially odd number (A268335).
Dirichlet g.f.: zeta(2*s) * zeta(2*s-2) * Product_{p prime} (1 + 1/p^s - 1/p^(3*s-2)).
Sum_{k=1..n} a(k) ~ c * n^(3/2), where c = (zeta(3)/3) * Product_{p prime} (1 + 1/p^(3/2) - 1/p^(5/2)) = 0.69451968056653021193... .

A383795 Dirichlet g.f.: zeta(2*s-2) * zeta(s)^2.

Original entry on oeis.org

1, 2, 2, 7, 2, 4, 2, 12, 12, 4, 2, 14, 2, 4, 4, 33, 2, 24, 2, 14, 4, 4, 2, 24, 28, 4, 22, 14, 2, 8, 2, 54, 4, 4, 4, 84, 2, 4, 4, 24, 2, 8, 2, 14, 24, 4, 2, 66, 52, 56, 4, 14, 2, 44, 4, 24, 4, 4, 2, 28, 2, 4, 24, 139, 4, 8, 2, 14, 4, 8, 2, 144, 2, 4, 56, 14, 4, 8, 2, 66, 113
Offset: 1

Views

Author

Vaclav Kotesovec, May 10 2025

Keywords

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[OddQ[e], Sum[(e+1-2*k) * p^(2*k), {k, 0, (e-1)/2}], Sum[(e+1-2*k) * p^(2*k), {k, 0, e/2}]]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, May 24 2025 *)
  • PARI
    for(n=1, 100, print1(direuler(p=2, n, 1/((1-p^2*X^2)*(1-X)^2))[n], ", "))

Formula

Sum_{k=1..n} a(k) ~ zeta(3/2)^2 * n^(3/2)/3 - n*(log(n) + 2*log(2*Pi) + 2*gamma - 1)/2, where gamma is the Euler-Mascheroni constant A001620.
Multiplicative with a(p^e) = Sum_{k=0..(e-1)/2} (e+1-2*k) * p^(2*k) if e is odd, and Sum_{k=0..e/2} (e+1-2*k) * p^(2*k) if e is even. - Amiram Eldar, May 24 2025
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