A218446 a(n) = Sum_{k>=0} floor(n/(5*k + 3)).
0, 0, 0, 1, 1, 1, 2, 2, 3, 4, 4, 4, 5, 6, 6, 7, 8, 8, 10, 10, 10, 11, 11, 12, 14, 14, 15, 16, 17, 17, 18, 18, 19, 21, 21, 21, 23, 23, 24, 26, 27, 27, 28, 29, 29, 30, 31, 31, 34, 34, 34, 35, 36, 37, 39, 39, 41, 42, 43, 43, 44, 44, 44, 46, 47, 48, 50, 50, 51, 53, 53, 53, 56, 57, 57, 58, 59, 59, 62, 62, 63
Offset: 0
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..10000
- 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.
Programs
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Mathematica
Accumulate[Table[Count[Divisors[n],?(Mod[#,5]==3&)],{n,0,90}]] (* _Harvey P. Dale, Nov 08 2012 *)
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Maxima
A218446[n]:=sum(floor(n/(5*k+3)),k,0,n)$ makelist(A218446[n],n,0,80); /* Martin Ettl, Oct 29 2012 */
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PARI
a(n)=sum(k=0,n,(n\(5*k+3)))
Formula
a(n) = n*log(n)/5 + c*n + O(n^(1/3)*log(n)), where c = gamma(3,5) - (1 - gamma)/5 = A256848 - (1 - A001620)/5 = -0.0983206... (Smith and Subbarao, 1981). - Amiram Eldar, Apr 20 2025