OFFSET
1,4
LINKS
Seiichi Manyama, Table of n, a(n) for n = 1..10000
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
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
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 *)
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Vladeta Jovovic, Nov 21 2002
STATUS
approved