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G.f.: Sum_{n>0} x^n/(1-x^(3n)) = Sum_{n>=0} x^(3n+1)/(1-x^(3n+1)).
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%I #51 Nov 25 2023 04:44:37

%S 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,

%T 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,

%U 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

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

%C 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

%D Bruce C. Berndt, On a certain theta-function in a letter of Ramanujan from Fitzroy House, Ganita 43 (1992), 33-43.

%H Nick Hobson, <a href="/A001817/b001817.txt">Table of n, a(n) for n = 1..10000</a>

%H P. G. Dirichlet, <a href="http://dx.doi.org/10.1515/crll.1840.21.1">Recherches sur diverses applications de l'analyse infinitésimale à la théorie des nombres</a>, J. Reine Angew. Math. 21 (1840), 1-12.

%H Michael Gilleland, <a href="/selfsimilar.html">Some Self-Similar Integer Sequences</a>.

%H R. A. Smith and M. V. Subbarao, <a href="https://doi.org/10.4153/CMB-1981-005-3">The average number of divisors in an arithmetic progression</a>, Canadian Mathematical Bulletin, Vol. 24, No. 1 (1981), pp. 37-41.

%F Moebius transform is period 3 sequence [1, 0, 0, ...]. - _Michael Somos_, Sep 20 2005

%F 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

%F Equals A051731 * [1, 0, 0, 1, 0, 0, 1, 0, 0, 1, ...]. - _Gary W. Adamson_, Nov 06 2007

%F a(n) = (A035191(n) + A002324(n)) / 2. - _Reinhard Zumkeller_, Nov 26 2011

%F 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

%e x + x^2 + x^3 + 2*x^4 + x^5 + x^6 + 2*x^7 + 2*x^8 + x^9 + ...

%p A001817 := proc(n)

%p local a,d ;

%p a := 0 ;

%p for d in numtheory[divisors](n) do

%p if modp(d,3) = 1 then

%p a := a+1 ;

%p end if ;

%p end do:

%p a ;

%p end proc:

%p seq(A001817(n),n=1..100) ; # _R. J. Mathar_, Sep 25 2017

%t a[n_] := DivisorSum[n, Boole[Mod[#, 3] == 1]&]; Array[a, 100] (* _Jean-François Alcover_, Dec 01 2015 *)

%o (PARI) a(n)=if(n<1,0,sumdiv(n,d,d%3==1))

%o (Haskell)

%o a001817 n = length [d | d <- [1,4..n], mod n d == 0]

%o -- _Reinhard Zumkeller_, Nov 26 2011

%Y Cf. A001822, A035191, A002324.

%Y Cf. A051731.

%Y Cf. A001620, A256425.

%K nonn,easy

%O 1,4

%A _N. J. A. Sloane_