A360658 -id:A360658 - cp's OEIS frontend

A360390 a(1) = 1; a(n) = -Sum_{k=2..n} k^2 * a(floor(n/k)).

1, -4, -13, -9, -34, 11, -38, -38, -38, 87, -34, -70, -239, 6, 231, 231, -58, -58, -419, -519, -78, 527, -2, -2, -2, 843, 843, 647, -194, -1319, -2280, -2280, -1191, 254, 1479, 1479, 110, 1915, 3436, 3436, 1755, -450, -2299, -2783, -2783, -138, -2347, -2347, -2347, -2347, 254, -422

Offset: 1

Seiichi Manyama, Apr 01 2023

sign

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

Partial sums of A359485.
Cf. A092149, A359478, A360658.
Cf. A336276.

f[p_, e_] := If[e == 1, -p^2, 0]; f[2, e_] := Switch[e, 1, -5, 2, 4, , 0]; s[1] = 1; s[n] := Times @@ f @@@ FactorInteger[n]; Accumulate[Array[s, 100]] (* Amiram Eldar, May 10 2023 *)

from functools import lru_cache
@lru_cache(maxsize=None)
def A360390(n):
    if n <= 1:
        return 1
    c, j = 0, 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c -= (j2*(j2-1)*((j2<<1)-1)-j*(j-1)*((j<<1)-1))//6*A360390(k1)
        j, k1 = j2, n//j2
    return c-(n*(n+1)*((n<<1)+1)-j*(j-1)*((j<<1)-1))//6 # Chai Wah Wu, Apr 01 2023

Sum_{k=1..n} k^2 * a(floor(n/k)) = 0 for n > 1.

G.f. A(x) satisfies x * (1 - x) = Sum_{k>=1} k^2 * (1 - x^k) * A(x^k).

A359478 a(1) = 1; a(n) = -Sum_{k=2..n} k * a(floor(n/k)).

Original entry on oeis.org

1, -2, -5, -3, -8, 1, -6, -6, -6, 9, -2, -8, -21, 0, 15, 15, -2, -2, -21, -31, -10, 23, 0, 0, 0, 39, 39, 25, -4, -49, -80, -80, -47, 4, 39, 39, 2, 59, 98, 98, 57, -6, -49, -71, -71, -2, -49, -49, -49, -49, 2, -24, -77, -77, -22, -22, 35, 122, 63, 93, 32, 125, 125, 125, 190, 91

Offset: 1

Seiichi Manyama, Mar 31 2023

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

Partial sums of A359484.
Cf. A092149, A360390, A360658.
Cf. A359479.

s[n_] := n * MoebiusMu[n] - If[OddQ[n], 0, MoebiusMu[n/2]*n/2]; Accumulate[Array[s, 100]] (* Amiram Eldar, May 09 2023 *)

from functools import lru_cache
@lru_cache(maxsize=None)
def A359478(n):
    if n <= 1:
        return 1
    c, j = 0, 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c -= (j2*(j2-1)-j*(j-1)>>1)*A359478(k1)
        j, k1 = j2, n//j2
    return c-(n*(n+1)-(j-1)*j>>1) # Chai Wah Wu, Mar 31 2023

Sum_{k=1..n} k * a(floor(n/k)) = 0 for n > 1.

G.f. A(x) satisfies x * (1 - x) = Sum_{k>=1} k * (1 - x^k) * A(x^k).

A359531 a(1) = 1, a(2) = -9; a(n) = -n^3 * Sum_{d|n, d < n} a(d) / d^3.

Original entry on oeis.org

1, -9, -27, 8, -125, 243, -343, 0, 0, 1125, -1331, -216, -2197, 3087, 3375, 0, -4913, 0, -6859, -1000, 9261, 11979, -12167, 0, 0, 19773, 0, -2744, -24389, -30375, -29791, 0, 35937, 44217, 42875, 0, -50653, 61731, 59319, 0, -68921, -83349, -79507, -10648, 0

Offset: 1

Seiichi Manyama, Apr 01 2023

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

Partial sums give A360658.
Cf. A092673, A359484, A359485.
Cf. A334659.

f[p_, e_] := If[e == 1, -p^3, 0]; f[2, e_] := Switch[e, 1, -9, 2, 8, , 0]; a[1] = 1; a[n] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, May 10 2023 *)

a(n) is multiplicative with a(2)= -9, a(4)= 8, a(2^e)= 0 if e>2. a(p)= -p^3, a(p^e)= 0 if e>1, p>2.

G.f. A(x) satisfies x * (1 - x) = Sum_{k>=1} k^3 * A(x^k).

cp's OEIS Frontend

A360390 a(1) = 1; a(n) = -Sum_{k=2..n} k^2 * a(floor(n/k)).

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A359478 a(1) = 1; a(n) = -Sum_{k=2..n} k * a(floor(n/k)).

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Author

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Links

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Programs

Mathematica

Python

Formula

A359531 a(1) = 1, a(2) = -9; a(n) = -n^3 * Sum_{d|n, d < n} a(d) / d^3.

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Author

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Links

Crossrefs

Programs

Mathematica

Formula