OFFSET
1,1
COMMENTS
The sequence is nondecreasing - this follows from the properties of the sum-of-divisors (sigma) and Euler's totient (phi) functions. Many terms appear more than once. Each integer greater than 73 appears at least once.
Colossally abundant (CA) numbers m are listed in A004490.
REFERENCES
G. H. Hardy and E.M. Wright, An introduction to the theory of numbers, 6th edition, Oxford University Press (2008), 350-353.
G. Robin, Grandes valeurs de la fonction somme des diviseurs et hypothèse de Riemann. J. Math. Pures Appl. 63 (1984), 187-213.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
L. Alaoglu and P. Erdos, On highly composite and similar numbers, Trans. Amer. Math. Soc., 56 (1944), 448-469. Errata
Keith Briggs, Abundant numbers and the Riemann Hypothesis, Experimental Math., Vol. 16 (2006), p. 251-256.
T. H. Grönwall, Some asymptotic expressions in the theory of numbers, Trans. Amer. Math. Soc 14 (1913), 113-122.
J.-L. Nicolas, Petites valeurs de la fonction d'Euler, J. Number Theory 17, no.3 (1983), 375-388.
S. Ramanujan, Highly composite numbers, Annotated and with a foreword by J.-L. Nicolas and G. Robin, Ramanujan J., 1 (1997), 119-153.
Eric W. Weisstein, MathWorld: Robin's Theorem
FORMULA
(1) sigma(m)/phi(m) ~ exp(2*gamma)*(log(log(m)))^2 as m tends to infinity.
Here gamma is the Euler constant, gamma = 0.5772156649...
Formula (1) can be proved based on two known facts for CA numbers m:
(A) sigma(m)/m ~ exp(gamma) * log(log(m)) [see Ramanujan, 1997, eq. 383]
(B) m/phi(m) ~ exp(gamma) * log(log(m))
(we get (1) simply by multiplying (A) and (B) together).
The following empirical inequality suggests that sigma(m)/phi(m) approximates the limiting sequence exp(2*gamma)*(log(log(m)))^2 from below:
(2) sigma(m)/phi(m) < exp(2*gamma)*(log(log(m)))^2 for large enough CA numbers m (namely, for m>10^35, i.e., beginning with the 34th CA number m). No formal proof is known for formula (2). If a proof of (2) becomes available, then Robin's inequality sigma(m)/m < exp(gamma) * log(log(m)) (and therefore the Riemann Hypothesis) will follow as well. Thus (2) must be exceedingly difficult to prove.
EXAMPLE
3 = [3/1] for m=2: sigma(2)=3 and phi(2)=1;
6 = [12/2] for m=6: sigma(6)=12 and phi(6)=2;
7 = [28/4] for m=12: sigma(12)=28 and phi(12)=4;
10 = [168/16] for m=60 (see A004490 for further values of m);
11 = [360/32]
12 = [1170/96]
16 = [9360/576]
16 = [19344/1152]
20 = [232128/11520]
23 = [3249792/138240]
23 = [6604416/276480]
24 = [20321280/829440]
25 = [104993280/4147200]
28 = [1889879040/66355200]
31 = [37797580800/1194393600]
34 = [907141939200/26276659200]
34 = [1828682956800/52553318400]
37 = [54860488704000/1471492915200]
39 = [1755535638528000/44144787456000]
40 = [12508191424512000/309013512192000]
40 = [37837279059148800/927040536576000]
43 = [1437816604247654400/33373459316736000]
CROSSREFS
KEYWORD
nonn
AUTHOR
Alexei Kourbatov, Feb 28 2012
STATUS
approved