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Let M(n) = maximal value of (n/k)^k over all k = 1, 2, ...; a(n) = floor(M(n)).
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%I #3 Mar 30 2012 16:50:49

%S 1,2,3,4,6,9,12,18,27,39,57,81,118,172,244,359,517,743,1085,1554,2254,

%T 3270,4667,6818,9846,14116,20589,29619,42762,62088,89055,129307,

%U 187064,267893,390499,563208,811020,1178088,1694774,2452059,3551313,5097655,7405861,10698505

%N Let M(n) = maximal value of (n/k)^k over all k = 1, 2, ...; a(n) = floor(M(n)).

%e The sequence of M(n)'s begins 1, 2, 3, 4, 6.2500000000000000000, 9, 12.703703703703703703..., 18.962962962962962963..., 27, 39.062500000000000000, 57.191406250000000000, 81, 118.81376000000000000, 172.10368000000000000, 244.14062500000000000, ...

%p Digits:=20; g:=(n,k)->evalf( (n/k)^k );

%p # for M(n):

%p f:=proc(n) local i,a; global g; a:=1; for i from 1 to 2*n do if g(n,i) > a then a:=g(n,i); fi; od: a; end;

%p # for A139076:

%p f1:=proc(n) local i,a; global g; a:=1; for i from 1 to 2*n do if g(n,i) > a then a:=g(n,i); fi; od: floor(a); end;

%p # for A139077:

%p f2:=proc(n) local i,a; global g; a:=1; for i from 1 to 2*n do if g(n,i) > a then a:=g(n,i); fi; od: round(a); end;

%p # for A139078:

%p f3:=proc(n) local i,a; global g; a:=1; for i from 1 to 2*n do if g(n,i) > a then a:=g(n,i); fi; od: ceil(a); end;

%Y Suggested by A000792. Cf. A139050, A139051, A139077, A139078.

%K nonn

%O 1,2

%A _N. J. A. Sloane_, Jun 03 2008