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.

A336278 a(n) = Sum_{k=1..n} mu(k)*k^4.

Original entry on oeis.org

1, -15, -96, -96, -721, 575, -1826, -1826, -1826, 8174, -6467, -6467, -35028, 3388, 54013, 54013, -29508, -29508, -159829, -159829, 34652, 268908, -10933, -10933, -10933, 446043, 446043, 446043, -261238, -1071238, -1994759, -1994759, -808838, 527498, 2028123
Offset: 1

Views

Author

Donald S. McDonald, Jul 15 2020

Keywords

Comments

Conjecture: a(n) changes sign infinitely often.

Crossrefs

Programs

  • Mathematica
    Array[Sum[MoebiusMu[k]*k^4, {k, #}] &, 35] (* Michael De Vlieger, Jul 15 2020 *)
    Accumulate[Table[MoebiusMu[x]x^4,{x,40}]] (* Harvey P. Dale, Jan 14 2021 *)
  • PARI
    a(n) = sum(k=1, n, moebius(k)*k^4); \\ Michel Marcus, Jul 15 2020
    
  • Python
    from functools import lru_cache
    @lru_cache(maxsize=None)
    def A336278(n):
        if n <= 1:
            return 1
        c, j = 1, 2
        k1 = n//j
        while k1 > 1:
            j2 = n//k1 + 1
            c -= (j2*(j2**2*(j2*(6*j2 - 15) + 10) - 1)-j*(j**2*(j*(6*j - 15) + 10) - 1))//30*A336278(k1)
            j, k1 = j2, n//j2
        return c-(n*(n**2*(n*(6*n + 15) + 10) - 1)-j*(j**2*(j*(6*j - 15) + 10) - 1))//30 # Chai Wah Wu, Apr 04 2023

Formula

Partial sums of A334660.
From Seiichi Manyama, Apr 03 2023: (Start)
G.f. A(x) satisfies x = Sum_{k>=1} k^4 * (1 - x^k) * A(x^k).
Sum_{k=1..n} k^4 * a(floor(n/k)) = 1. (End)

A336279 a(n) = Sum_{k=1..n} mu(k)*k^5.

Original entry on oeis.org

1, -31, -274, -274, -3399, 4377, -12430, -12430, -12430, 87570, -73481, -73481, -444774, 93050, 852425, 852425, -567432, -567432, -3043531, -3043531, 1040570, 6194202, -242141, -242141, -242141, 11639235, 11639235, 11639235, -8871914, -33171914, -61801065
Offset: 1

Views

Author

Donald S. McDonald, Jul 15 2020

Keywords

Comments

Conjecture: a(n) changes sign infinitely often.

Crossrefs

Programs

  • Mathematica
    Array[Sum[MoebiusMu[k]*k^5, {k, #}] &, 32] (* Michael De Vlieger, Jul 15 2020 *)
  • PARI
    a(n) = sum(k=1, n, moebius(k)*k^5); \\ Michel Marcus, Jul 15 2020
    
  • Python
    from functools import lru_cache
    @lru_cache(maxsize=None)
    def A336279(n):
        if n <= 1:
            return 1
        c, j = 1, 2
        k1 = n//j
        while k1 > 1:
            j2 = n//k1 + 1
            c -= (j2**2*(j2**2*(j2*(2*j2 - 6) + 5) - 1)-j**2*(j**2*(j*(2*j - 6) + 5) - 1))//12*A336279(k1)
            j, k1 = j2, n//j2
        return c-(n**2*(n**2*(n*(2*n + 6) + 5) - 1)-j**2*(j**2*(j*(2*j - 6) + 5) - 1))//12 # Chai Wah Wu, Apr 04 2023

Formula

From Seiichi Manyama, Apr 03 2023: (Start)
G.f. A(x) satisfies x = Sum_{k>=1} k^5 * (1 - x^k) * A(x^k).
Sum_{k=1..n} k^5 * a(floor(n/k)) = 1. (End)

A344430 a(n) = Sum_{k=1..n} mu(k) * k^k.

Original entry on oeis.org

1, -3, -30, -30, -3155, 43501, -780042, -780042, -780042, 9999219958, -275312450653, -275312450653, -303150419042906, 10808856406515110, 448702746787374485, 448702746787374485, -826791559139549389692, -826791559139549389692
Offset: 1

Views

Author

Seiichi Manyama, May 19 2021

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := Sum[MoebiusMu[k] * k^k, {k,1,n}]; Array[a, 20] (* Amiram Eldar, May 19 2021 *)
    Accumulate[Table[MoebiusMu[n]n^n,{n,20}]] (* Harvey P. Dale, Jan 25 2022 *)
  • PARI
    a(n) = sum(k=1, n, moebius(k)*k^k);
    
  • Python
    from sympy import mobius
    def A344430(n): return sum(mobius(k)*k**k for k in range(1,n+1)) # Chai Wah Wu, Apr 05 2023
Showing 1-3 of 3 results.