A279060 Number of divisors of n of the form 6*k + 1.
0, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 1, 3, 2, 1, 2, 1, 1, 2, 2, 2, 1, 1, 1, 2, 2, 2, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 4, 1, 2, 1, 2, 1, 2, 3, 1, 2, 1, 1, 2, 2, 2, 1, 1, 1, 2, 2, 2, 2, 1, 2, 2, 1, 2, 1, 2, 1
Offset: 0
Examples
a(14) = 2 because 14 has 4 divisors {1,2,7,14} among which 2 divisors {1,7} are of the form 6*k + 1.
Links
- Antti Karttunen, Table of n, a(n) for n = 0..65537
- 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.
Crossrefs
Programs
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Mathematica
nmax = 120; CoefficientList[Series[Sum[x^k/(1 - x^(6 k)), {k, 1, nmax}], {x, 0, nmax}], x] nmax = 120; CoefficientList[Series[Sum[x^(6 k + 1)/(1 - x^(6 k + 1)), {k, 0, nmax}], {x, 0, nmax}], x] Table[Count[Divisors[n],?(Mod[#,6]==1&)],{n,0,120}] (* _Harvey P. Dale, Apr 27 2018 *) -
PARI
A279060(n) = if(!n,n,sumdiv(n, d, (1==(d%6)))); \\ Antti Karttunen, Jul 09 2017
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Python
from sympy import divisors def A279060(n): return sum(d%6 == 1 for d in divisors(n)) # David Radcliffe, Jun 19 2025
Formula
G.f.: Sum_{k>=1} x^k/(1 - x^(6*k)).
G.f.: Sum_{k>=0} x^(6*k+1)/(1 - x^(6*k+1)).
From Antti Karttunen, Oct 03 2018: (Start)
a(n) = A320001(n) + [1 == n (mod 6)], where [ ] is the Iverson bracket, giving 1 only when n = 1 mod 6, and 0 otherwise.
Sum_{k=1..n} a(k) = n*log(n)/6 + c*n + O(n^(1/3)*log(n)), where c = gamma(1,6) - (1 - gamma)/6 = 0.686263..., gamma(1,6) = -(psi(1/6) + log(6))/6 is a generalized Euler constant, and gamma is Euler's constant (A001620) (Smith and Subbarao, 1981). - Amiram Eldar, Nov 25 2023
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