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 1-2 of 2 results.

A364268 a(n) = Sum_{k=1..n} k^2*sigma_2(k), where sigma_2 is A001157.

Original entry on oeis.org

1, 21, 111, 447, 1097, 2897, 5347, 10787, 18158, 31158, 45920, 76160, 104890, 153890, 212390, 299686, 383496, 530916, 661598, 879998, 1100498, 1395738, 1676108, 2165708, 2572583, 3147183, 3744963, 4568163, 5276285, 6446285, 7370767, 8768527, 10097107
Offset: 1

Views

Author

Seiichi Manyama, Oct 20 2023

Keywords

Crossrefs

Programs

  • Mathematica
    Accumulate[Table[n^2*DivisorSigma[2, n], {n, 1, 33}]] (* Amiram Eldar, Oct 20 2023 *)
  • PARI
    f(n, m) = (subst(bernpol(m+1, x), x, n+1)-subst(bernpol(m+1, x), x, 0))/(m+1);
    a(n, s=2, t=2) = sum(k=1, n, k^(s+t)*f(n\k, s));
    
  • Python
    def A364268(n): return sum(k**4*(m:=n//k)*(m+1)*((m<<1)+1)//6 for k in range(1,n+1)) # Chai Wah Wu, Oct 20 2023
    
  • Python
    from math import isqrt
    def A364268(n): return (((s:=isqrt(n))*(s+1)*(2*s+1))**2*(1-3*s*(s+1))//6 + sum((q:=n//k)*(q+1)*(2*q+1)*k**2*(5*k**2+3*q*(q+1)-1) for k in range(1,s+1)))//30 # Chai Wah Wu, Oct 21 2023

Formula

a(n) = Sum_{k=1..n} k^4 * A000330(floor(n/k)).
a(n) ~ (zeta(3)/5) * n^5. - Amiram Eldar, Oct 20 2023

A364194 a(n) = Sum_{k=1..n} k^3*sigma(k), where sigma is A000203.

Original entry on oeis.org

1, 25, 133, 581, 1331, 3923, 6667, 14347, 23824, 41824, 57796, 106180, 136938, 202794, 283794, 410770, 499204, 726652, 863832, 1199832, 1496184, 1879512, 2171520, 3000960, 3485335, 4223527, 5010847, 6240159, 6971829, 8915829, 9869141, 11933525, 13658501
Offset: 1

Views

Author

Seiichi Manyama, Oct 20 2023

Keywords

Crossrefs

Partial sums of A282211.

Programs

  • Mathematica
    Accumulate[Table[n^3*DivisorSigma[1, n], {n, 1, 33}]] (* Amiram Eldar, Oct 20 2023 *)
  • PARI
    f(n, m) = (subst(bernpol(m+1, x), x, n+1)-subst(bernpol(m+1, x), x, 0))/(m+1);
    a(n, s=3, t=1) = sum(k=1, n, k^(s+t)*f(n\k, s));
    
  • Python
    def A364194(n): return sum((k**2*(m:=n//k)*(m+1)>>1)**2 for k in range(1,n+1)) # Chai Wah Wu, Oct 20 2023
    
  • Python
    from math import isqrt
    def A364194(n): return ((((s:=isqrt(n))*(s + 1))**3*(2*s+1)*(1-3*s*(s+1))>>1) + sum((q:=n//k)*(q+1)*k**3*(q*(15*k+q*(15*k+12*q+18)+2)-2) for k in range(1,s+1)))//60 # Chai Wah Wu, Oct 21 2023

Formula

a(n) = Sum_{k=1..n} k^4 * A000537(floor(n/k)).
a(n) ~ (zeta(2)/5) * n^5. - Amiram Eldar, Oct 20 2023
Showing 1-2 of 2 results.