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a(n) = Sum_{k=0..n} floor(n/(3*k + 1)).
2

%I #21 Nov 25 2023 08:03:08

%S 0,1,2,3,5,6,7,9,11,12,14,15,17,19,21,22,25,26,27,29,32,34,36,37,39,

%T 41,43,44,48,49,51,53,56,57,59,61,63,65,67,69,73,74,76,78,81,82,84,85,

%U 88,91,94,95,99,100,101,103,107,109,111,112,115,117,119,121,125,127,129,131,134,135,139,140,142,144,146,148,152

%N a(n) = Sum_{k=0..n} floor(n/(3*k + 1)).

%H Amiram Eldar, <a href="/A218442/b218442.txt">Table of n, a(n) for n = 0..10000</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 a(n) = 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

%t d[n_] := DivisorSum[n, 1 &, Mod[#, 3] == 1 &]; d[0] = 0; Accumulate@Array[d, 100, 0] (* _Amiram Eldar_, Nov 25 2023 *)

%o (PARI) a(n)=sum(k=0,n\3,(n\(3*k+1)))

%o (Maxima) A218442[n]:=sum(floor(n/(3*k+1)),k,0,n)$

%o makelist(A218442[n],n,0,80); /* _Martin Ettl_, Oct 29 2012 */

%Y Partial sums of A001817.

%Y Cf. A001620, A218443, A256425.

%K nonn,easy

%O 0,3

%A _Benoit Cloitre_, Oct 28 2012

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