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A001817
G.f.: Sum_{n>0} x^n/(1-x^(3n)) = Sum_{n>=0} x^(3n+1)/(1-x^(3n+1)).
22
1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 1, 2, 3, 2, 2, 1, 2, 2, 2, 1, 4, 1, 2, 2, 3, 1, 2, 2, 2, 2, 2, 2, 4, 1, 2, 2, 3, 1, 2, 1, 3, 3, 3, 1, 4, 1, 1, 2, 4, 2, 2, 1, 3, 2, 2, 2, 4, 2, 2, 2, 3, 1, 4, 1, 2, 2, 2, 2, 4, 2, 2, 2, 5, 1, 2, 1, 4, 2, 2, 1, 4, 1, 2, 4, 3, 2, 2, 2, 3, 2, 3, 1, 5, 1, 2, 2, 4, 2
OFFSET
1,4
COMMENTS
a(n) is the number of positive divisors of n of the form 3k+1. If r(n) denotes the number of representations of n by the quadratic form j^2+ij+i^2, then r(n)= 6 *(a(n)-A001822(n)). - Benoit Cloitre, Jun 24 2002
REFERENCES
Bruce C. Berndt, On a certain theta-function in a letter of Ramanujan from Fitzroy House, Ganita 43 (1992), 33-43.
LINKS
R. A. Smith and M. V. Subbarao, The average number of divisors in an arithmetic progression, Canadian Mathematical Bulletin, Vol. 24, No. 1 (1981), pp. 37-41.
FORMULA
Moebius transform is period 3 sequence [1, 0, 0, ...]. - Michael Somos, Sep 20 2005
G.f.: Sum_{k>0} x^(3k-2)/(1-x^(3k-2)) = Sum_{k>0} x^k/(1-x^(3k)). - Michael Somos, Sep 20 2005
Equals A051731 * [1, 0, 0, 1, 0, 0, 1, 0, 0, 1, ...]. - Gary W. Adamson, Nov 06 2007
a(n) = (A035191(n) + A002324(n)) / 2. - Reinhard Zumkeller, Nov 26 2011
Sum_{k=1..n} a(k) = n*log(n)/3 + c*n + O(n^(1/3)*log(n)), where c = gamma(1,3) - (1 - gamma)/3 = A256425 - (1 - A001620)/3 = 0.536879... (Smith and Subbarao, 1981). - Amiram Eldar, Nov 25 2023
EXAMPLE
x + x^2 + x^3 + 2*x^4 + x^5 + x^6 + 2*x^7 + 2*x^8 + x^9 + ...
MAPLE
A001817 := proc(n)
local a, d ;
a := 0 ;
for d in numtheory[divisors](n) do
if modp(d, 3) = 1 then
a := a+1 ;
end if ;
end do:
a ;
end proc:
seq(A001817(n), n=1..100) ; # R. J. Mathar, Sep 25 2017
MATHEMATICA
a[n_] := DivisorSum[n, Boole[Mod[#, 3] == 1]&]; Array[a, 100] (* Jean-François Alcover, Dec 01 2015 *)
PROG
(PARI) a(n)=if(n<1, 0, sumdiv(n, d, d%3==1))
(Haskell)
a001817 n = length [d | d <- [1, 4..n], mod n d == 0]
-- Reinhard Zumkeller, Nov 26 2011
CROSSREFS
KEYWORD
nonn,easy
STATUS
approved