A078181 a(n) = Sum_{d|n, d == 1 (mod 3)} d.

1, 1, 1, 5, 1, 1, 8, 5, 1, 11, 1, 5, 14, 8, 1, 21, 1, 1, 20, 15, 8, 23, 1, 5, 26, 14, 1, 40, 1, 11, 32, 21, 1, 35, 8, 5, 38, 20, 14, 55, 1, 8, 44, 27, 1, 47, 1, 21, 57, 36, 1, 70, 1, 1, 56, 40, 20, 59, 1, 15, 62, 32, 8, 85, 14, 23, 68, 39, 1, 88, 1, 5, 74, 38, 26, 100, 8, 14, 80, 71, 1

Offset: 1

Author

Vladeta Jovovic, Nov 21 2002

Keywords

• easy
• nonn

Links

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

Crossrefs

Cf. A001817, A001822, A035382, A051731, A272716, A353908.

Cf. Sum_{d|n, d==1 mod k} d: A000593 (k=2), this sequence (k=3), A050449 (k=4), A284097 (k=5), A284098 (k=6), A284099 (k=7), A284100 (k=8).

Programs

Maple

A078181 := proc(n)
    a := 0 ;
    for d in numtheory[divisors](n) do
        if modp(d,3) =1 then
            a :=a+d ;
        end if;
    end do:
    a;
end proc: # R. J. Mathar, May 11 2016

Mathematica

a[n_] := Plus @@ Select[Divisors[n], Mod[#, 3] == 1 &]; Array[a, 100] (* Giovanni Resta, May 11 2016 *)

Formula

G.f.: Sum_{n>=0} (3*n+1)*x^(3*n+1)/(1-x^(3*n+1)).

G.f.: -q*P'/P where P = Product_{n>=0} (1 - q^(3*n+1)). - Joerg Arndt, Aug 03 2011

Conjecture. If a(n)=n+1 then n==1 (mod 3). (Is this easy to settle? It has been verified for n=1,2,3,...,2000.) - John W. Layman, Apr 03 2006

The conjecture is false. The first and only counterexample below 10^8 is a(6800) = 6801 and 6800 == 2 (mod 3). - Lambert Herrgesell (zero815(AT)googlemail.com), May 06 2008

Equals A051731 * [1, 0, 0, 4, 0, 0, 7, 0, 0, 10, ...]. - Gary W. Adamson, Nov 06 2007

A272027(n/3) + a(n) + A078182(n) = A000203(n). - R. J. Mathar, May 25 2020

G.f.: Sum_{n >= 1} x^n*(1 + 2*x^(3*n))/(1 - x^(3*n))^2. - Peter Bala, Dec 19 2021

Sum_{k=1..n} a(k) = c * n^2 + O(n*log(n)), where c = Pi^2/36 = 0.274155... (A353908). - Amiram Eldar, Nov 26 2023