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Least m > 0 such that 2*m^n <= Sum_{k < m} k^n.
1

%I #21 Jun 23 2020 06:22:46

%S 3,5,8,10,13,15,18,20,23,25,28,30,33,35,38,40,42,45,47,50,52,55,57,60,

%T 62,65,67,70,72,75,77,79,82,84,87,89,92,94,97,99,102,104,107,109,112,

%U 114,116,119,121,124,126,129,131,134,136,139,141,144,146,149,151,153,156,158,161,163,166

%N Least m > 0 such that 2*m^n <= Sum_{k < m} k^n.

%C Obviously a(n) is a lower limit for any s solution to 2*s^n = Sum_{x in S} x^n, S subset of {1, ..., s-1}.

%C First differences are (2, 3, 2, 3, ...) except for a duplicated 2 in positions {16, 31, 46, 61, 76, 91; 104, 119, 134, 149, 164, 179, 194, 209, 224, 239, 254, 269; 282, 297, ...}: Here the first differences are always 15 except for a 13 after the 6th, 18th, ... term.

%F a(n) >= A195168(n+1) with equality for n not in {13, 15; 26, 28, 30; 39, 41, 43, 45; 52, 54, ..., 60; 65, 67, ..., 75, 78, 80, ..., 90; 89, 91, ..., 103; 102, 104, ..., 114, 115, ...} \ {120, 122, 124, 126, 135, 137, 139, 150, 152, 165}.

%F a(n) <= A047218(n+2) with equality for n <= 17 and even n <= 34.

%F Conjecture: a(n) = round(n/log(3/2) + 3).

%e For n=0, 2*m^0 = 2 > Sum_{k<m} k^0 = m - 1 <=> 3 > m, so a(0) = 3.

%e For n=1, 2*m^1 > Sum_{k<m} k^1 = m(m-1)/2 <=> 4 > m - 1, so a(1) = 5.

%t Table[Block[{m = 1, s = 0}, While[2 m^n > s, s = s + m^n; m++]; m], {n, 0, 66}] (* _Michael De Vlieger_, Apr 30 2020 *)

%o (PARI) apply( A332102(n, s)=for(m=1, oo, s<2*m^n||return(m); s+=m^n), [0..66])

%Y Cf. A332101 (same without factor 2 in definition).

%Y Cf. A195168, A047218, A029919 (all have common initial terms but differ later and only remain lower resp. upper bounds).

%K nonn

%O 0,1

%A _M. F. Hasler_, Apr 18 2020