A180638 Phan Thành Nam's upper bound on the number of non-relativistic electrons bound to a nucleus of charge n.
4, 6, 8, 9, 11, 12, 14, 15, 17, 18, 20, 21, 23, 24, 25, 27, 28, 30, 31, 32, 34, 35, 36, 38, 39, 40, 42, 43, 44, 46, 47, 48, 50, 51, 52, 54, 55, 56, 58, 59, 60, 62, 63, 64, 65, 67, 68, 69, 71, 72, 73, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 89, 90, 91, 93, 94, 95, 96, 98, 99, 100, 102
Offset: 1
Examples
a(5) = floor(1.22*5 + 3.1*5^(1/3)) = floor(11.40...) = 11, which means that boron (the element with atomic number 5, i.e., with 5 protons) can have no more than 11 bound electrons, which would give it a -6 charge. B^5- has been observed (in Al_3BC) so this bound is reasonably tight. - _Charles R Greathouse IV_, Sep 14 2016
References
- E. H. Lieb and R. Seiringer, The stability of matter in quantum mechanics, Cambridge University Press, 2009.
Links
- E. H. Lieb, Bound on the maximum negative ionization of atoms and molecules, Phys. Rev. A 29, 3018-3028 (1984).
- Phan Thành Nam, New bounds on the maximum ionization of atoms , Communications in Mathematical Physics 312:2 (2012), pp. 427-445. arXiv:1009.2367 [math-ph]
- L. A. Seco, I. M. Sigal, and J. P. Solovej, Bound on the ionization energy of large atoms, Communications in Mathematical Physics 131:2 (1990), pp. 307-315.
Crossrefs
Cf. A007656.
Programs
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Maple
A180638 := proc(n) 1.22*n+3.1*n^(1/3) ; floor(%) ; end proc: seq(A180638(n),n=1..100) ; # R. J. Mathar, Oct 06 2010
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Mathematica
Table[Floor[1.22 n + 3.1 n^(1/3)], {n, 73}] (* Michael De Vlieger, Dec 23 2015 *)
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PARI
a(n)=floor(1.22*n+3.1*sqrtn(n,3)) \\ Charles R Greathouse IV, Sep 13 2016
Formula
a(n) = floor(1.22*n + 3.1*n^(1/3)).
Extensions
Reference converted to link, entries checked - R. J. Mathar, Oct 06 2010
Comments