

A062011


Number of cyclic subgroups of the group C_n X C_2 (where C_n is the cyclic group with n elements).


12



2, 4, 4, 6, 4, 8, 4, 8, 6, 8, 4, 12, 4, 8, 8, 10, 4, 12, 4, 12, 8, 8, 4, 16, 6, 8, 8, 12, 4, 16, 4, 12, 8, 8, 8, 18, 4, 8, 8, 16, 4, 16, 4, 12, 12, 8, 4, 20, 6, 12, 8, 12, 4, 16, 8, 16, 8, 8, 4, 24, 4, 8, 12, 14, 8, 16, 4, 12, 8, 16, 4, 24, 4, 8, 12, 12, 8, 16, 4, 20, 10, 8, 4, 24, 8, 8, 8, 16
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OFFSET

1,1


COMMENTS

More generally, the number of cyclic subgroups of the group C_n X C_m is Sum_{in, jm} phi(i)*phi(j)/phi(lcm(i,j)), where phi=Euler totient function, cf. A000010.  Vladeta Jovovic, Jul 15 2001
Also number of divisors of p*n, where p is any prime not dividing n, e.g.: a(n) = A000005(A087560(n)) = A000005(A119416(n)).  Reinhard Zumkeller, May 17 2006
If p(x) is a polynomial with integer coefficients, and if r is an integer zero of p(x), then r is a divisor of the constant term c_0 of p(x). Under this theorem, p(x) can have a(c_0) possible integer roots. a(n) is also the number of integer divisor of n, while A000005(n) is the number of positive divisors.  Enrique Pérez Herrero, Jul 21 2011


LINKS

Harry J. Smith, Table of n, a(n) for n = 1..1000
Omar E. Pol, Illustration of initial terms
Index entries for sequences related to groups


FORMULA

a(n) = 2*tau(n) = 2*A000005(n).
L.g.f.: log(Product_{k>=1} (1  x^k)^(2/k)) = Sum_{n>=1} a(n)*x^n/n.  Ilya Gutkovskiy, Mar 18 2018


MATHEMATICA

A062011[n_] := 2*DivisorSigma[0, n]; Array[A062011, 50] (* Enrique Pérez Herrero, Jul 15 2011 *)


PROG

(PARI) for (n=1, 1000, write("b062011.txt", n, " ", 2*numdiv(n))) \\ Harry J. Smith, Jul 29 2009


CROSSREFS

Cf. A000005, A060648, A060710.
Sequence in context: A049782 A091666 A084290 * A225520 A132857 A152782
Adjacent sequences: A062008 A062009 A062010 * A062012 A062013 A062014


KEYWORD

nonn,easy


AUTHOR

Ahmed Fares (ahmedfares(AT)mydeja.com), Jul 12 2001


EXTENSIONS

More terms from Vladeta Jovovic, Jul 14 2001


STATUS

approved



