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127 (just those submitted by Marot) |
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Just showing those entries submitted by 'Marot': (Click here to show all) 
127 = 10^2+3^3 = gives 127 = 2^2*5^2+3^3 where p(1) appears three times, p(2) twice and p(3) once. 127 = 2^6+3^3+6^2 (palindromic) = p(31) = p(p(11)) = p(p(p(5))) = p(p(p(p(3)))) = p(p(p(p(p(2))))) = p(p(p(p(p(p(1)))))). 127 is the prime for which ((p(1) + ... + p(n))/p(n)) decreases for the first time. [Marot]
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