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A001876
Number of divisors of n of the form 5k+1; a(0)=0.
18
0, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 1, 2, 1, 1, 2, 2, 1, 2, 1, 2, 1, 1, 1, 2, 2, 2, 2, 1, 1, 3, 1, 1, 1, 1, 2, 3, 1, 2, 1, 2, 1, 3, 1, 1, 2, 2, 1, 2, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 1, 4, 1, 1, 1, 1, 2, 3, 1, 1, 1, 2, 2, 3, 1, 2, 2, 2, 1, 3, 1, 2, 1
OFFSET
0,7
LINKS
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.
FORMULA
G.f.: Sum_{n>=0} x^(5n+1)/(1-x^(5n+1)).
G.f.: Sum_{n>=1} x^n/(1-x^(5*n)). - Joerg Arndt, Jan 30 2011
Sum_{k=1..n} a(k) = 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, Nov 25 2023
MATHEMATICA
CoefficientList[ Series[ Together[ Sum[ x^n/(1 - x^(5n)), {n, 110}]], {x, 0, 110}], x] (* Robert G. Wilson v, Jan 31 2011 *)
a[n_] := DivisorSum[n, 1 &, Mod[#, 5] == 1 &]; a[0] = 0; Array[a, 100, 0] (* Amiram Eldar, Nov 25 2023 *)
PROG
(PARI) a(n) = if(n==0, 0, sumdiv(n, d, (d % 5) == 1)); \\ Michel Marcus, Feb 25 2021
CROSSREFS
For numbers of divisors of n of the form 5k+i (i=1, 2, 3, 4) see: this sequence, A001877, A001878, A001899.
Sequence in context: A215870 A097587 A001179 * A033182 A053797 A254011
KEYWORD
nonn,easy
STATUS
approved