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

%I #2 Mar 30 2012 17:30:25

%S 1,2,3,4,6,7,9,11,12,14,16,18,20,23,25,27,29,32,34,37,39,42,44,47,49,

%T 52,55,57,60,63,66,69,72,75,77,80,83,86,90,93,96,99,102,105,108,112,

%U 115,118,122,125,128,132,135,138,142,145,149,152,156,159,163,167,170

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

%t s = 0; k = 1; Do[ While[ s <= n, s = s + N[ 1/k^(1/4), 24 ]; k++ ]; Print[ k - 1 ], {n, 1, 75} ]

%Y Cf. A019529.

%K nonn

%O 0,2

%A _Robert G. Wilson v_, Aug 01 2000