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.

A259445 Multiplicative with a(n) = n if n is odd and a(2^s)=2.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 2, 9, 10, 11, 6, 13, 14, 15, 2, 17, 18, 19, 10, 21, 22, 23, 6, 25, 26, 27, 14, 29, 30, 31, 2, 33, 34, 35, 18, 37, 38, 39, 10, 41, 42, 43, 22, 45, 46, 47, 6, 49, 50, 51, 26, 53, 54, 55, 14, 57, 58, 59, 30, 61, 62, 63, 2, 65, 66, 67, 34
Offset: 1

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Comments

If n = 2^s*m with m odd and s > 0 then a(n) = 2*m.

Crossrefs

Programs

  • Maple
    A259445 := proc(n::integer)
        local a, pe, p,e ;
        a := 1 ;
        for pe in ifactors(n)[2] do
            p := op(1,pe) ;
            e := op(2,pe) ;
            if p = 2 then
                a := 2*a ;
            else
                a := a*p^e  ;
            end if;
        end do:
        a;
    end proc:
    seq(A259445(n),n=1..80) ; # R. J. Mathar, Feb 21 2025
  • Mathematica
    G[n_] := If[Mod[n, 2] == 0, n/2^(FactorInteger[n][[1, 2]] - 1), n]; Table[G[n], {n, 1, 70}]
  • PARI
    a(n)=n>>max(valuation(n,2)-1,0) \\ Charles R Greathouse IV, Jun 28 2015

Formula

From Peter Bala, Feb 21 2019: (Start)
a(n) = n*gcd(n,2)/gcd(n,2^n).
a(2*n) = 2*A000265(2*n); a(2*n+1) = A000265(2*n+1).
O.g.f.: x*(1 + 4*x + x^2)/(1 - x^2)^2 - 2*( F(x^2) + F(x^4) + F(x^8) + ... ), where F(x) = x/(1 - x)^2.
O.g.f. for reciprocals: Sum_{n >= 1} (1/a(n))*x^n = (3/4)*L(x) - (1/4)*L(-x) + (1/4)*( L(x^2) + L(x^4) + L(x^8) + ... ), where L(x) = log(1/(1 - x)).
(End)
From Peter Bala, Mar 09 2019: (Start)
a(n) = (-1)^(n+1)*Sum_ {d divides n} (-1)^(d+n/d)*phi(d), where phi(n) = A000010(n) is the Euler totient function. Cf. the identity n = Sum_ {d divides n} phi(d). Cf. A046897 and A321558.
O.g.f.: Sum_{n >= 1} phi(n)*x^n/(1 + (-x)^n). (End)
From Amiram Eldar, Nov 28 2022: (Start)
Dirichlet g.f.: zeta(s-1)*(1 + 1/2^(s-1) - 2/(2^s-1)).
Sum_{k=1..n} a(k) ~ (5/12) * n^2. (End)
a(n) = n /A160467(n). - R. J. Mathar, Feb 21 2025