A218445 -id:A218445 - cp's OEIS frontend

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

Benoit Cloitre, Oct 28 2012

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.

Partial sums of A001878.
Cf. A218444, A218445, A218447.
Cf. A001620, A256848.

Accumulate[Table[Count[Divisors[n],?(Mod[#,5]==3&)],{n,0,90}]] (* _Harvey P. Dale, Nov 08 2012 *)

A218446[n]:=sum(floor(n/(5*k+3)),k,0,n)$
makelist(A218446[n],n,0,80); /* Martin Ettl, Oct 29 2012 */

PARI
```
a(n)=sum(k=0,n,(n\(5*k+3)))
```

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

A218444 a(n) = Sum_{k>=0} floor(n/(5*k + 1)).

Original entry on oeis.org

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

Benoit Cloitre, Oct 28 2012

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.

Partial sums of A001876.
Cf. A218444, A218445, A218446.
Cf. A001620, A256779.

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 *)

A218444[n]:=sum(floor(n/(5*k+1)),k,0,n)$
makelist(A218444[n],n,0,80); /* Martin Ettl, Oct 29 2012 */

PARI
```
a(n)=sum(k=0,n,(n\(5*k+1)))
```

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

A218447 a(n) = Sum_{k>=0} floor(n/(5*k + 4)).

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 1, 1, 2, 3, 3, 3, 4, 4, 5, 5, 6, 6, 7, 8, 9, 9, 9, 9, 11, 11, 11, 12, 14, 15, 15, 15, 16, 16, 17, 17, 19, 19, 20, 21, 22, 22, 23, 23, 25, 26, 26, 26, 28, 29, 29, 29, 30, 30, 32, 32, 34, 35, 36, 37, 38, 38, 38, 39, 41, 41, 41, 41, 43, 44, 45, 45, 48, 48, 49, 49, 51, 51, 52, 53, 54

Offset: 0

Benoit Cloitre, Oct 28 2012

Robert Israel, 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.

Partial sums of A001899.
Cf. A218444, A218445, A218446.
Cf. A001620, A256849.

g:= n -> nops(select(t -> t mod 5 = 4, numtheory:-divisors(n))):
g(0):= 0:
ListTools:-PartialSums(map(g, [$0..100])); # Robert Israel, Apr 29 2021

A218447[n]:=sum(floor(n/(5*k+4)),k,0,n)$
makelist(A218447[n],n,0,80); /* Martin Ettl, Oct 20 2012 */

PARI
```
a(n)=sum(k=0,n,(n\(5*k+4)))
```

a(n) = n*log(n)/5 + c*n + O(n^(1/3)*log(n)), where c = gamma(4,5) - (1 - gamma)/5 = A256849 - (1 - A001620)/5 = -0.213442... (Smith and Subbarao, 1981). - Amiram Eldar, Apr 20 2025

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A218446 a(n) = Sum_{k>=0} floor(n/(5*k + 3)).

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A218444 a(n) = Sum_{k>=0} floor(n/(5*k + 1)).

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A218447 a(n) = Sum_{k>=0} floor(n/(5*k + 4)).

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