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.

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

Views

Author

Seiichi Manyama, Mar 31 2023

Keywords

Crossrefs

Partial sums of A359484.
Cf. A359479.

Programs

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

Formula

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).

A361985 a(1) = 1, a(2) = 1; a(n) = n * Sum_{d|n, d < n} (-1)^(n/d) a(d) / d.

Original entry on oeis.org

1, 1, -3, 6, -5, -3, -7, 24, 0, -5, -11, -18, -13, -7, 15, 96, -17, 0, -19, -30, 21, -11, -23, -72, 0, -13, 0, -42, -29, 15, -31, 384, 33, -17, 35, 0, -37, -19, 39, -120, -41, 21, -43, -66, 0, -23, -47, -288, 0, 0, 51, -78, -53, 0, 55, -168, 57, -29, -59, 90, -61, -31, 0, 1536, 65, 33, -67, -102, 69, 35, -71, 0, -73, -37, 0
Offset: 1

Views

Author

Seiichi Manyama, Apr 02 2023

Keywords

Crossrefs

Partial sums give A359479.
Cf. A359484.

Programs

  • Mathematica
    f[p_, e_] := If[e == 1, -p, 0]; f[2, e_] := If[e == 1, 1, 6*4^(e-2)]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, May 09 2023 *)

Formula

a(n) is multiplicative with a(2) = 1, a(2^e) = 6*4^(e-2) if e>1. a(p) = -p, a(p^e) = 0 if e>1, p>2.
G.f. A(x) satisfies -x * (1 - x) = Sum_{k>=1} (-1)^k * k * A(x^k).

A361981 a(1) = 1; a(n) = Sum_{k=2..n} (-1)^k * k^2 * a(floor(n/k)).

Original entry on oeis.org

1, 4, -5, 23, -2, -29, -78, 146, 146, 71, -50, -302, -471, -618, -393, 1399, 1110, 1110, 749, 49, 490, 127, -402, -2418, -2418, -2925, -2925, -4297, -5138, -4463, -5424, 8912, 10001, 9134, 10359, 10359, 8990, 7907, 9428, 3828, 2147, 3470, 1621, -1767, -1767, -3354, -5563
Offset: 1

Views

Author

Seiichi Manyama, Apr 02 2023

Keywords

Crossrefs

Partial sums of A361986.

Programs

  • Mathematica
    f[p_, e_] := If[e == 1, -p^2, 0]; f[2, e_] := If[e == 1, 3, 7*2^(3*e-4)]; s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; Accumulate[Array[s, 100]] (* Amiram Eldar, May 09 2023 *)
  • Python
    from functools import lru_cache
    @lru_cache(maxsize=None)
    def A361981(n):
        if n <= 1:
            return 1
        c, j = 0, 2
        k1 = n//j
        while k1 > 1:
            j2 = n//k1 + 1
            c += ((j2*(j2-1) if j2&1 else -j2*(j2-1))+(-j*(j-1) if j&1 else j*(j-1))>>1)*A361981(k1)
            j, k1 = j2, n//j2
        return c+((-n*(n+1) if n&1 else n*(n+1))+(-j*(j-1) if j&1 else j*(j-1))>>1) # Chai Wah Wu, Apr 02 2023

Formula

Sum_{k=1..n} (-1)^k * k^2 * a(floor(n/k)) = 0 for n > 1.
G.f. A(x) satisfies -x * (1 - x) = Sum_{k>=1} (-1)^k * k^2 * (1 - x^k) * A(x^k).
Showing 1-3 of 3 results.