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 1-3 of 3 results.

A077455 a(n) = sigma_4(n^4)/sigma(n^4).

Original entry on oeis.org

1, 2255, 360205, 8965359, 195688121, 812262275, 11869610005, 36654862063, 190649623129, 441276712855, 2853329308061, 3229367138595, 21506735660905, 26765970561275, 70487839624805, 150121132912367, 548357292625505, 429914900155895, 2096841596815405, 1754414256800439
Offset: 1

Views

Author

Benoit Cloitre, Nov 30 2002

Keywords

Examples

			a(2) = sigma_4(2^4)/sigma(2^4) = 69905/31 = 2255.
		

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := (p^(12*e+3) + p^(8*e+2) + p^(4*e+1) + 1)/(p^3 + p^2 + p + 1); a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 20] (* Amiram Eldar, Sep 09 2020 *)
  • PARI
    a(n)=sumdiv(n^4,d,d^4)/sigma(n^4)
    
  • PARI
    a(n) = my(f=factor(n^4)); sigma(f, 4)/sigma(f); \\ Michel Marcus, Sep 09 2020

Formula

a(n) = A001158(n^4)/A000203(n^4).
Multiplicative with a(p^e) = (p^(12*e+3) + p^(8*e+2) + p^(4*e+1) + 1)/(p^3 + p^2 + p + 1). - Amiram Eldar, Sep 09 2020
Sum_{k=1..n} a(k) ~ c * n^13, where c = (zeta(3)*zeta(5)*zeta(9)*zeta(13)/13) * Product_{p prime} (1-1/p^2-1/p^3+1/p^5-1/p^7+1/p^8-1/p^12+2/p^13-2/p^14+2/p^15-1/p^16+2/p^17-3/p^18+1/p^19+1/p^21-1/p^22-1/p^26-1/p^27) = 0.048281563902... . - Amiram Eldar, Nov 20 2022

A077456 a(n) = sigma_5(n^5)/sigma(n^5).

Original entry on oeis.org

1, 549791, 2337334621, 567767102431, 76323251878121, 1285045538614211, 68398022066406901, 595065340418751455, 8138648440293876241, 41961836973324022711, 611595047235520833101, 1327061705176829563651, 17543094367661056941241, 37604616949911916507691
Offset: 1

Views

Author

Benoit Cloitre, Nov 30 2002

Keywords

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := (p^(20*e+4) + p^(15*e+3) + p^(10*e+2) + p^(5*e+1) + 1)/(p^4 + p^3 + p^2 + p + 1); a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 20] (* Amiram Eldar, Sep 09 2020 *)
    Table[DivisorSigma[5,n^5]/DivisorSigma[1,n^5],{n,20}] (* Harvey P. Dale, Mar 05 2022 *)
  • PARI
    a(n)=sumdiv(n^5,d,d^5)/sigma(n^5)
    
  • PARI
    a(n) = my(f=factor(n^5)); sigma(f, 5)/sigma(f); \\ Michel Marcus, Sep 09 2020

Formula

a(n) = A001160(n^5)/A000203(n^5).
Multiplicative with a(p^e) = (p^(20*e+4) + p^(15*e+3) + p^(10*e+2) + p^(5*e+1) + 1)/(p^4 + p^3 + p^2 + p + 1). - Amiram Eldar, Sep 09 2020

A077457 a(n) = sigma_4(n^4)/sigma_2(n^4).

Original entry on oeis.org

1, 205, 5905, 52429, 375601, 1210525, 5649505, 13421773, 38742049, 76998205, 212601841, 309593245, 810932305, 1158148525, 2217923905, 3435973837, 6951703105, 7942120045, 16936647121, 19692384829, 33360327025, 43583377405, 78163228705, 79255569565, 146719125601
Offset: 1

Views

Author

Benoit Cloitre, Nov 30 2002

Keywords

Comments

sigma_y(n^x) divides sigma_x(n^x) for all n if y divides x.

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := (p^(8*e+2) + 1)/(p^2 + 1); a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 25] (* Amiram Eldar, Sep 09 2020 *)
  • PARI
    a(n)=sumdiv(n^4,d,d^4)/sumdiv(n^4,d,d^2)
    
  • PARI
    a(n) = my(f=factor(n^4)); sigma(f, 4)/sigma(f, 2); \\ Michel Marcus, Sep 09 2020

Formula

a(n) = A001159(n^4)/A001157(n^4).
Multiplicative with a(p^e) = (p^(8*e+2) + 1)/(p^2 + 1). - Amiram Eldar, Sep 09 2020
Sum_{k=1..n} a(k) ~ c * n^9, where c = (zeta(9)/9) * Product_{p prime} (1 - 1/p^3 + 1/p^5 - 1/p^7) = 0.09549806119... . - Amiram Eldar, Oct 28 2022
Showing 1-3 of 3 results.