I see this formula in a book, it comes from R.L.Graham,AMM(1995) : $$\frac{1}{3}=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{7^2}+\frac{1}{54^2}+\frac{1}{112^2}+\frac{1}{640^2}+\frac{1}{4032^2}+\frac{1}{10080^2}+\frac{1}{24192^2}+\frac{1}{40320^2}+\frac{1}{120960^2}$$ But I can not find this article in AMM(1955),can you help me to find which month it belongs to ?Thank you!

Addition:There is another example in Paul Erdos and Egyptian Fractions:



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And in [38]http://www.math.ucsd.edu/~ronspubs/64_07_reciprocals.pdf , I find a better formula: $$\frac{1}{3}=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{20^2}+\frac{1}{30^2}+\frac{1}{60^2}$$

$PS$: Since I have found the above papers,this problem has been less important.Thanks for your attention!

  • $\begingroup$ His papers are listed on math.ucsd.edu/~ronspubs and the only 1995 paper in the American Mathematical Monthly does not seem to contain this formula. $\endgroup$ – TMM May 10 '13 at 12:07
  • $\begingroup$ @TMM,thank you,I have found it before..And I have dowloaded all the AMM at 1955,but I didn't find his paper when I searched for his name. $\endgroup$ – Next May 10 '13 at 12:19
  • $\begingroup$ Is it 1995 or 1955?! $\endgroup$ – DonAntonio May 10 '13 at 14:58
  • $\begingroup$ @DonAntonio,the book says it's 1955. $\endgroup$ – Next May 10 '13 at 15:06
  • 1
    $\begingroup$ @Hecke What book is this from? It may be possible to reverse engineer the right citation. $\endgroup$ – Andrés E. Caicedo May 10 '13 at 15:18

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