A formula by R.L.Graham,AMM(1995)

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:

http://www.math.ucsd.edu/~ronspubs/pre_Egyptian.pdf

$$\frac{1}{2}=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{15^2}+\frac{1}{16^2}+\frac{1}{36^2}+\frac{1}{60^2}+\frac{1}{180^2}$$

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!

• 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. – TMM May 10 '13 at 12:07
• @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. – Next May 10 '13 at 12:19
• Is it 1995 or 1955?! – DonAntonio May 10 '13 at 14:58
• @DonAntonio,the book says it's 1955. – Next May 10 '13 at 15:06
• @Hecke What book is this from? It may be possible to reverse engineer the right citation. – Andrés E. Caicedo May 10 '13 at 15:18