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Integer averages of two successive perfect powers (pp(n) + pp(n+1))/2.
1

%I #4 Mar 30 2012 17:26:03

%S 6,26,34,123,136,206,234,352,498,1012,1350,1746,2082,2192,2203,2724,

%T 3075,3428,4977,5804,6874,7760,8050,8146,9335,10732,12244,13874,16724,

%U 17500,19782,21928,24519,26948,29860,32946,35829,39254,42862,50639

%N Integer averages of two successive perfect powers (pp(n) + pp(n+1))/2.

%e 6 is OK because (pp(2) + pp(3))/2 = (4 + 8)/2 = 6.

%K easy,nonn

%O 1,1

%A _Zak Seidov_, Oct 11 2002