

A018227


Magic numbers: atoms with full shells containing any of these numbers of electrons are considered electronically stable.


21



2, 10, 18, 36, 54, 86, 118, 168, 218, 290, 362, 460, 558, 686, 814, 976, 1138, 1338, 1538, 1780, 2022, 2310, 2598, 2936, 3274, 3666, 4058, 4508, 4958, 5470, 5982, 6560, 7138, 7786, 8434, 9156, 9878, 10678, 11478, 12360, 13242, 14210, 15178
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OFFSET

1,1


COMMENTS

Atomic numbers of noble elements in the periodic table.
Comment from Don N. Page (don(AT)phys.ualberta.ca), Dec 12 2006: (Start)
"Relativistic corrections and instabilities to pair creation of electrons and positrons would occur even if one could have stable nuclei of arbitrarily many protons Z for the fixed value of the fine structure constant alpha ~ 1/137 in our universe.
"However, if one considered an imaginary universe with arbitrarily tiny alpha and a fixed point source of charge Z, one could have stable neutral atoms of Z nonrelativistic electrons of mass m for any Z, so long as one takes the limit Z alpha > 0 by taking alpha > 0 after fixing Z.
"One could then define noble elements to be given by the integer values of Z such that the ionization energy, in units of m c^2 alpha^2, of any such atom in its ground state with larger Z is less than that of the noble element (which appears to be the case for all the noble elements with the actual nonzero value of alpha).
"This sequence of idealized nonrelativistic noble elements with Z electrons would give an infinite sequence of integers Z, which may or may not be the same as that given by the explicit formula listed for the present sequence. It would likely be a difficult mathematical problem to calculate this infinite sequence." (End)


LINKS



FORMULA

a(n) = a(n1) + ((2*n + 3 + (1)^n)^2)/8; a(n) = (2*n^3 + 12*n^2 + 25*n  6 + (1)^n*(3*n + 6))/12.  Warut Roonguthai, Jun 20 2005
a(n) = n{(n+3)^2 + 5}/6 for even n, a(n) = n{(n+3)^2 + 2)/6  1 [or C(n+3,3)  2, i.e., A000292(n)  2] for odd n.  Lekraj Beedassy, Feb 02 2006
a(n) = n*((n+3)^2 + 2)/6 + (n+2)*(1+(1)^n)/4  1, n >= 1.
a(n) = (n+1)*(n+2)*(n+3)/6 + (n+2)*(1+(1)^n)/4  2, n >= 1.
a(n) = T_{n+1} + (n+2)*(1+(1)^n)/4  2, n >= 1, where T_n is the nth tetrahedral number.
G.f.: 2*x*(1 + 3*x  2*x^2  x^3 + x^4)/((1  x)^4*(1 + x)^2). (End)


PROG

(Magma) [n*((n+3)^2 + 2)/6 + (n+2)*(1+(1)^n)/4  1: n in [1..50]]; // Vincenzo Librandi, May 03 2011


CROSSREFS

Cf. A018226 for the magic numbers for nucleons (protons and neutrons).


KEYWORD

nonn,easy


AUTHOR

John Raithel (raithel(AT)rahul.net)


STATUS

approved



