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.

A309127 a(n) = n + 2^4 * floor(n/2^4) + 3^4 * floor(n/3^4) + 4^4 * floor(n/4^4) + ...

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139
Offset: 1

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Author

Ilya Gutkovskiy, Jul 13 2019

Keywords

Comments

Partial sums of A300909.

Crossrefs

Programs

  • Mathematica
    Table[Sum[k^4 Floor[n/k^4], {k, 1, n}], {n, 1, 75}]
    nmax = 75; CoefficientList[Series[1/(1 - x) Sum[k^4 x^(k^4)/(1 - x^(k^4)), {k, 1, Floor[nmax^(1/4)] + 1}], {x, 0, nmax}], x] // Rest
  • PARI
    a(n) = sum(k=1, n, k^4*(n\k^4)); \\ Seiichi Manyama, Aug 30 2021

Formula

G.f.: (1/(1 - x)) * Sum_{k>=1} k^4 * x^(k^4)/(1 - x^(k^4)).
a(n) ~ zeta(5/4)*n^(5/4)/5 - n/2. - Vaclav Kotesovec, Aug 30 2021

A333844 G.f.: Sum_{k>=1} k * x^(k^4) / (1 - x^(k^4)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3
Offset: 1

Views

Author

Ilya Gutkovskiy, Apr 07 2020

Keywords

Comments

Sum of 4th roots of 4th powers dividing n.

Crossrefs

Programs

  • Mathematica
    nmax = 112; CoefficientList[Series[Sum[k x^(k^4)/(1 - x^(k^4)), {k, 1, Floor[nmax^(1/4)] + 1}], {x, 0, nmax}], x] // Rest
    Table[DivisorSum[n, #^(1/4) &, IntegerQ[#^(1/4)] &], {n, 112}]
    f[p_, e_] := (p^(Floor[e/4] + 1) - 1)/(p - 1); a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Dec 01 2020 *)

Formula

Dirichlet g.f.: zeta(s) * zeta(4*s-1).
If n = Product (p_j^k_j) then a(n) = Product ((p_j^(floor(k_j/4) + 1) - 1)/(p_j - 1)).
Sum_{k=1..n} a(k) ~ zeta(3)*n + zeta(1/2)*sqrt(n)/2. - Vaclav Kotesovec, Dec 01 2020

A347398 Expansion of g.f. Sum_{k>=1} k^k * x^(k^k)/(1 - x^(k^k)).

Original entry on oeis.org

1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 28, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 28, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 28, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 32, 1, 1, 1, 5
Offset: 1

Views

Author

Seiichi Manyama, Aug 30 2021

Keywords

Examples

			1^1 | 108, 2^2 | 108 and 3^3 | 108. So a(108) = 1^1 + 2^2 + 3^3 = 32.
		

Crossrefs

Programs

  • PARI
    a(n) = sum(k=1, n, (n%k^k==0)*k^k);

Formula

a(n) = A347397(n) - A347397(n-1) for n > 1.
a(n) = Sum_{k=1..n, k^k | n} k^k.
Showing 1-3 of 3 results.