A218444 a(n) = Sum_{k>=0} floor(n/(5*k + 1)).
0, 1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 15, 16, 17, 18, 20, 21, 23, 24, 25, 27, 29, 30, 32, 33, 35, 36, 37, 38, 40, 42, 44, 46, 47, 48, 51, 52, 53, 54, 55, 57, 60, 61, 63, 64, 66, 67, 70, 71, 72, 74, 76, 77, 79, 81, 83, 84, 85, 86, 88, 90, 92, 94, 96, 97, 101, 102, 103, 104, 105
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
a[n_] := Sum[ Floor[n/(5*k+1)], {k, 0, Ceiling[n/5]}]; Table[a[n], {n, 0, 70}] (* Jean-François Alcover, Feb 22 2013 *) -
Maxima
A218444[n]:=sum(floor(n/(5*k+1)),k,0,n)$ makelist(A218444[n],n,0,80); /* Martin Ettl, Oct 29 2012 */
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PARI
a(n)=sum(k=0,n,(n\(5*k+1)))
Formula
a(n) = Sum_{k>=0} floor(n/(5*k + 1)).
a(n) = n*log(n)/5 + c*n + O(n^(1/3)*log(n)), where c = gamma(1,5) - (1 - gamma)/5 = A256779 - (1 - A001620)/5 = 0.651363... (Smith and Subbarao, 1981). - Amiram Eldar, Apr 20 2025