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This \u201ccomplexity\u201d turns out to be approximately 3.75 \u00d7 10<jats:sup>20<\/jats:sup> and it is found necessary to invoke the device of modulo arithmetic and the \u201cChinese Remainder Theorem\u201d in order to evaluate it precisely on a small computer. The exact spanningtree count for buckminsterfullerene is 375 291 866 372 898 816 000, or, 2<jats:sup>25<\/jats:sup> \u00d7 3<jats:sup>4<\/jats:sup> \u00d7 5<jats:sup>3<\/jats:sup> \u00d7 11<jats:sup>5<\/jats:sup> \u00d7 19<jats:sup>3<\/jats:sup>. A \u201cringcurrent\u201d calculation by the method of McWeeny may be based on any desired one of this vast number of spanning trees.<\/jats:p>","DOI":"10.1002\/jcc.540120909","type":"journal-article","created":{"date-parts":[[2005,1,2]],"date-time":"2005-01-02T00:38:21Z","timestamp":1104626301000},"page":"1118-1124","source":"Crossref","is-referenced-by-count":15,"title":["The number of spanning trees in buckminsterfullerene"],"prefix":"10.1002","volume":"12","author":[{"given":"T. J. N.","family":"Brown","sequence":"first","affiliation":[]},{"given":"R. B.","family":"Mallion","sequence":"additional","affiliation":[]},{"given":"P.","family":"Pollak","sequence":"additional","affiliation":[]},{"given":"Branca R. M.","family":"de Castro","sequence":"additional","affiliation":[]},{"given":"J. A. N. 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April 14 1990."},{"key":"e_1_2_1_44_2","unstructured":"We noted with some amusement (though we do not claim it to have any particular significance!) that when the complexity of icosahedral C60is expressed in the form\\documentclass{article}\\pagestyle{empty}\\begin{document}$$ 2^{25} \\times 3^4 \\times 5^3 \\times 11^5 \\times 19^3 $$\\end{document}the sum of the bases (2 + 3 + 5 + 11 + 19) is equal to the sum of the powers (25 + 4 + 3 + 5 + 3) both being 40."},{"key":"e_1_2_1_45_2","doi-asserted-by":"publisher","DOI":"10.1098\/rspa.1947.0102"},{"key":"e_1_2_1_46_2","doi-asserted-by":"publisher","DOI":"10.1246\/bcsj.63.765"},{"key":"e_1_2_1_47_2","unstructured":"This could also be looked at in the following way which may be intuitively rather more obvious: IfF V andEbe the number of faces vertices and edges respectively Euler's relation is\\documentclass{article}\\pagestyle{empty}\\begin{document}$$ F + V = E + 2 $$\\end{document}and so \\documentclass{article}\\pagestyle{empty}\\begin{document}$$ E - (F - 1) = V - 1. $$\\end{document}It is evident that any spanning tree (which contains allVvertices of the network but no circuits) must comprizeV\u2010 1 edges; an upper bound to the number of spanning trees is therefore ECV\u20101which by the above amounts toECE\u2010(F\u20101)=ECF\u20101. 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