A218445 a(n) = Sum_{k>=0} floor(n/(5*k + 2)).
0, 0, 1, 1, 2, 2, 3, 4, 5, 5, 6, 6, 8, 8, 10, 10, 11, 12, 13, 13, 14, 15, 17, 17, 19, 19, 20, 21, 23, 23, 24, 24, 26, 26, 28, 29, 31, 32, 33, 33, 34, 34, 37, 37, 39, 39, 40, 41, 43, 44, 45, 46, 48, 48, 50, 50, 52, 53, 54, 54, 56, 56, 58, 59, 61, 61, 63, 64, 66, 66, 68, 68, 71, 71, 73, 73, 74, 76, 77, 77, 78
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
Table[Sum[Floor[n/(5k+2)],{k,0,n}],{n,0,80}] (* Harvey P. Dale, Dec 08 2022 *) -
Maxima
A218445[n]:=sum(floor(n/(5*k+2)),k,0,n)$ makelist(A218445[n],n,0,80); /* Martin Ettl, Oct 29 2012 */
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PARI
a(n)=sum(k=0,n\5,(n\(5*k+2)))
Formula
a(n) = n*log(n)/5 + c*n + O(n^(1/3)*log(n)), where c = gamma(2,5) - (1 - gamma)/5 = A256780 - (1 - A001620)/5 = 0.105832... (Smith and Subbarao, 1981). - Amiram Eldar, Apr 20 2025