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a(1) = 1; a(m+1) = sum_{k=1 to m} [max(m, a(k))].
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%I #6 Apr 09 2014 10:15:49

%S 1,1,4,10,22,47,97,197,397,797,1597,3198,6400,12804,25612,51228,

%T 102460,204924,409852,819708,1639420,3278844,6557692,13115389,

%U 26230783,52461571,104923147,209846299,419692603,839385211,1678770427

%N a(1) = 1; a(m+1) = sum_{k=1 to m} [max(m, a(k))].

%e a(5) = max(4,1) + max(4,1) + max(4,4) + max(4,10) = 4 + 4 + 4 + 10 = 22.

%K easy,nonn

%O 1,3

%A _Leroy Quet_, Aug 04 2000