A344597 -id:A344597 - cp's OEIS frontend

A344596 a(n) = Sum_{k=1..n} mu(k) * (floor(n/k)^3 - floor((n-1)/k)^3).

1, 6, 18, 30, 60, 66, 126, 132, 198, 204, 330, 276, 468, 414, 552, 552, 816, 630, 1026, 840, 1116, 1050, 1518, 1128, 1740, 1476, 1890, 1692, 2436, 1704, 2790, 2256, 2820, 2544, 3384, 2556, 3996, 3186, 3960, 3408, 4920, 3420, 5418, 4260, 5112, 4686, 6486, 4560, 6930, 5340, 6816

Offset: 1

Seiichi Manyama, May 24 2021

nonn

Seiichi Manyama, Table of n, a(n) for n = 1..10000

Essentially 6*A102309 and 6*A326419.

Cf. A000578, A071778, A140434, A344597.

a[n_] := Sum[MoebiusMu[k] * First @ Differences @ (Quotient[{n - 1, n}, k]^3), {k, 1, n}]; Array[a, 50] (* Amiram Eldar, May 24 2021 *)

a(n) = sum(k=1, n, moebius(k)*((n\k)^3-((n-1)\k)^3));

a(n) = if(n<2, n, 3*sumdiv(n, d, moebius(n/d)*(d-1)*d));

my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, moebius(k)*x^k*(1+4*x^k+x^(2*k))/(1-x^k)^3))

my(N=66, x='x+O('x^N)); Vec(x+6*sum(k=1, N, moebius(k)*x^(2*k)/(1-x^k)^3))

from sympy import mobius, divisors
def A344596(n): return 3*sum(mobius(n//d)*d*(d-1) for d in divisors(n,generator=True)) if n>1 else 1 # Chai Wah Wu, May 09 2025

Sum_{k=1..n} a(k) * floor(n/k) = n^3.
Sum_{k=1..n} a(k) = A071778(n).
a(n) = 3 * Sum_{d|n} mu(n/d) * (d-1) * d for n > 1.
G.f.: Sum_{k >= 1} mu(k) * x^k * (1 + 4*x^k + x^(2*k))/(1 - x^k)^3.
G.f.: x + 6 * Sum_{k>=1} mu(k) * x^(2*k)/(1 - x^k)^3.
a(2^k) = 3*2^(k-2)*(3*2^k-2) for k>0. - Chai Wah Wu, May 10 2025

A383783 a(n) = Sum_{k=1..2^n} mu(k) * (floor(2^n/k)^4 - floor((2^n-1)/k)^4).

Original entry on oeis.org

1, 14, 160, 1520, 13216, 110144, 899200, 7266560, 58425856, 468583424, 3753379840, 30045900800, 240442679296, 1923843375104, 15391954862080, 123140470538240, 985143091265536, 7881222038749184, 63050085546065920, 504401921315962880, 4035220318323736576

Offset: 0

Chai Wah Wu, May 09 2025

nonn

Chai Wah Wu, Table of n, a(n) for n = 0..48

Cf. A082540, A344597.

a[n_]:=Sum[MoebiusMu[k]*(Floor[2^n/k]^4-Floor[(2^n-1)/k]^4),{k,2^n}]; Array[a,21,0] (* James C. McMahon, May 10 2025 *)

a(n) = sum(k=1, 2^n, moebius(k) * ((2^n\k)^4 - ((2^n-1)\k)^4)); \\ Michel Marcus, May 10 2025

from functools import lru_cache
@lru_cache(maxsize=None)
def A082540(n):
    if n == 0:
        return 0
    c, j = 1, 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c += (j2-j)*A082540(k1)
        j, k1 = j2, n//j2
    return n*(n**3-1)-c+j
def A383783(n): return A082540(m:=1<A082540(m-1)

a(n) = A344597(2^n) = A082540(2^n) - A082540(2^n-1).

cp's OEIS Frontend

A344596 a(n) = Sum_{k=1..n} mu(k) * (floor(n/k)^3 - floor((n-1)/k)^3).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

PARI

PARI

PARI

Python

Formula

A383783 a(n) = Sum_{k=1..2^n} mu(k) * (floor(2^n/k)^4 - floor((2^n-1)/k)^4).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Python

Formula