A094472 a(n) = n*tau(n) - sigma(n) - phi(n), where tau(n) is the number of divisors of n.
-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
Examples
As tau(10)= 4, sigma(10) = 18, phi(10) = 4, then a(10) = 10*4-18-4 = 18. - _Bernard Schott_, Feb 06 2020
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
- Michael De Vlieger, Table of n, a(n) for n = 1..10000
- A. Makowski, Aufgaben 339, Elemente der Mathematik 15 (1960), pp. 39-40.
Crossrefs
Programs
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Mathematica
Table[w*DivisorSigma[0, w]-DivisorSigma[1, w]-EulerPhi[w], {w, 1, 100}]
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PARI
apply( {A094472(n)=n*numdiv(n=factor(n))-sigma(n)-eulerphi(n)}, [1..99]) \\ M. F. Hasler, Feb 07 2020
Formula
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
Comments