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Sum of a(n) terms of 1/k^(6/7) first exceeds n.
0

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

%S 1,2,4,8,16,31,56,96,158,253,393,594,878,1271,1806,2523,3472,4711,

%T 6312,8359,10949,14199,18243,23237,29360,36816,45841,56698,69689,

%U 85152,103467,125060,150406,180034,214529,254542,300788,354056,415215,485213

%N Sum of a(n) terms of 1/k^(6/7) first exceeds n.

%e For example, a(4) = 16 since Sum_{k=1..16} 1/k^(6/7) = 4.014698427... > 4, whereas Sum_{k=1..15} 1/k^(6/7) = 3.921823784... < 4.

%t s = 0; k = 1; Do[ While[ s <= n, s = s + N[ 1/k^(7/8), 24 ]; k++ ]; Print[ k - 1 ], {n, 1, 40} ]

%Y Cf. A019529 and A002387.

%K nonn

%O 0,2

%A _Robert G. Wilson v_, Aug 01 2000