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A094472
a(n) = n*tau(n) - sigma(n) - phi(n), where tau(n) is the number of divisors of n.
1
-1, 0, 0, 3, 0, 10, 0, 13, 8, 18, 0, 40, 0, 26, 28, 41, 0, 63, 0, 70, 40, 42, 0, 124, 24, 50, 50, 100, 0, 160, 0, 113, 64, 66, 68, 221, 0, 74, 76, 214, 0, 228, 0, 160, 168, 90, 0, 340, 48, 187, 100, 190, 0, 294, 108, 304, 112, 114, 0, 536, 0, 122, 238, 289, 128, 364, 0, 250, 136, 392, 0, 645, 0, 146, 286, 280, 152, 432, 0, 582
OFFSET
1,4
COMMENTS
If n is prime, then a(n) = 0.
Is the reverse statement true [namely (a(n)=0 -> n=prime)]?
From Bernard Schott, Feb 06 2020: (Start)
The answer to this question is yes: a(n) = 0 iff n is prime (see the reference De Koninck & Mercier, Problème 625). This property comes from the 2 results below:
1) If f and g are multiplicative functions with positive values, then, for n >= 2 Sum_{d|n} f(d)*g(n/d) >= f(n) + g(n) with equality iff n is prime (see reference Problème 624).
2) Sum_{d|n} sigma(d)*phi(n/d) = n * tau(n) (see reference Problème 596).
Together, these 2 results give n * tau(n) >= sigma(n) + phi(n) with equality iff n is prime.
Also a(n) >= 0 for n > 1. (End)
REFERENCES
Jean-Marie De Koninck and Armel Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 625 pp. 82, 281; Problème 596, pp. 80, 275; Problème 624, pp. 82, 281; Ellipses, Paris, 2004.
LINKS
A. Makowski, Aufgaben 339, Elemente der Mathematik 15 (1960), pp. 39-40.
FORMULA
a(n) = n*A000005(n) - A000203(n) - A000010(n).
Sum_{k=1..n} a(k) ~ n^2*log(n)/2 + (gamma - 1/4 - Pi^2/12 - 3/Pi^2)*n^2, where gamma is Euler's constant (A001620). - Amiram Eldar, Dec 07 2023
EXAMPLE
As tau(10)= 4, sigma(10) = 18, phi(10) = 4, then a(10) = 10*4-18-4 = 18. - Bernard Schott, Feb 06 2020
MATHEMATICA
Table[w*DivisorSigma[0, w]-DivisorSigma[1, w]-EulerPhi[w], {w, 1, 100}]
PROG
(PARI) apply( {A094472(n)=n*numdiv(n=factor(n))-sigma(n)-eulerphi(n)}, [1..99]) \\ M. F. Hasler, Feb 07 2020
CROSSREFS
Cf. A000005 (tau), A000010 (phi), A000203 (sigma).
Cf. A038040 (n*tau(n)), A094471 (n*tau(n)-sigma(n)), A065387 (phi(n)+sigma(n)).
Sequence in context: A294106 A377268 A375991 * A347999 A028850 A138364
KEYWORD
sign,easy
AUTHOR
Labos Elemer, May 28 2004
STATUS
approved