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Sum of a(n) terms of 1/k^(4/5) first exceeds n.
0

%I #4 Sep 23 2016 16:10:57

%S 1,2,4,7,14,24,40,63,95,140,201,281,384,516,682,888,1141,1449,1820,

%T 2263,2789,3408,4133,4976,5951,7074,8360,9826,11492,13376,15499,17884,

%U 20554,23533,26849,30528,34600,39095,44045,49485,55450,61976,69103

%N Sum of a(n) terms of 1/k^(4/5) first exceeds n.

%t s = 0; k = 1; Do[ While[ s <= n, s = s + N[ 1/k^(4/5), 24 ]; k++ ]; Print[ k - 1 ], {n, 1, 50} ]

%t Join[{1},Table[Position[Accumulate[Table[N[1/k^(4/5)],{k,100000}]],_?(#>n&),{1},1],{n,50}]//Flatten] (* _Harvey P. Dale_, Sep 23 2016 *)

%Y Cf. A019529 and A002387.

%K nonn

%O 0,2

%A _Robert G. Wilson v_, Aug 01 2000