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.

A326575 Expansion of Sum_{k>=1} k * x^k * (1 + x^(2*k)) / (1 + x^(2*k) + x^(4*k)).

Original entry on oeis.org

1, 2, 3, 4, 4, 6, 8, 8, 9, 8, 10, 12, 14, 16, 12, 16, 16, 18, 20, 16, 24, 20, 22, 24, 21, 28, 27, 32, 28, 24, 32, 32, 30, 32, 32, 36, 38, 40, 42, 32, 40, 48, 44, 40, 36, 44, 46, 48, 57, 42, 48, 56, 52, 54, 40, 64, 60, 56, 58, 48, 62, 64, 72, 64, 56, 60
Offset: 1

Views

Author

Ilya Gutkovskiy, Sep 12 2019

Keywords

Examples

			G.f. = x + 2*x^2 + 3*x^3 + 4*x^4 + 4*x^5 + 6*x^6 + 8*x^7 + 8*x^8 + ... - _Michael Somos_, Oct 23 2019
		

Crossrefs

Cf. A003586 (fixed points), A035178, A050469, A122373, A326401.

Programs

  • Mathematica
    nmax = 66; CoefficientList[Series[Sum[k x^k (1 + x^(2 k))/(1 + x^(2 k) + x^(4 k)), {k, 1, nmax}], {x, 0, nmax}], x] // Rest
    Table[DivisorSum[n, # &, MemberQ[{1}, Mod[n/#, 6]] &] - DivisorSum[n, # &, MemberQ[{5}, Mod[n/#, 6]] &], {n, 1, 66}]
    f[p_, e_] := Which[p < 5, p^e, Mod[p, 6] == 5, (p^(e + 1) - (-1)^(e + 1))/(p + 1), Mod[p, 6] == 1, (p^(e + 1) - 1)/(p - 1)]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Dec 02 2020 *)
  • PARI
    a(n) = { sumdiv(n, d, d*((n/d%6==1)-(n/d%6==5))) } \\ Andrew Howroyd, Sep 12 2019
    
  • PARI
    {a(n) = if( n<0, 0, sumdiv( n, d, n/d * kronecker( -12, d)))}; /* Michael Somos, Oct 23 2019 */

Formula

a(n) = Sum_{d|n, n/d==1 (mod 6)} d - Sum_{d|n, n/d==5 (mod 6)} d.
G.f.: Sum_{k>=0} x^(6*k+1) / (1 - x^(6*k+1))^2 - x^(6*k+5) / (1 - x^(6*k+5))^2. - Michael Somos, Oct 23 2019
Multiplicative with a(p^e) = p^e if p < 5, (p^(e+1)-(-1)^(e+1))/(p+1) if p == 5 (mod 6), and (p^(e+1)-1)/(p-1) if p == 1 (mod 6). - Amiram Eldar, Dec 02 2020
Sum_{k=1..n} a(k) ~ c * n^2, where c = (1/2) * Product_{primes p == 5 (mod 6)} 1/(1+1/p^2) * Product_{primes p == 1 (mod 3)} 1/(1 - 1/p^2) = A340578 * A175646 / 2 = 0.48831400806... . - Amiram Eldar, Nov 06 2022