

A332102


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


1



3, 5, 8, 10, 13, 15, 18, 20, 23, 25, 28, 30, 33, 35, 38, 40, 42, 45, 47, 50, 52, 55, 57, 60, 62, 65, 67, 70, 72, 75, 77, 79, 82, 84, 87, 89, 92, 94, 97, 99, 102, 104, 107, 109, 112, 114, 116, 119, 121, 124, 126, 129, 131, 134, 136, 139, 141, 144, 146, 149, 151, 153, 156, 158, 161, 163, 166
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OFFSET

0,1


COMMENTS

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, ..., s1}.
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.


LINKS



FORMULA

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}.
a(n) <= A047218(n+2) with equality for n <= 17 and even n <= 34.
Conjecture: a(n) = round(n/log(3/2) + 3).


EXAMPLE

For n=0, 2*m^0 = 2 > Sum_{k<m} k^0 = m  1 <=> 3 > m, so a(0) = 3.
For n=1, 2*m^1 > Sum_{k<m} k^1 = m(m1)/2 <=> 4 > m  1, so a(1) = 5.


MATHEMATICA

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 *)


PROG

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


CROSSREFS

Cf. A332101 (same without factor 2 in definition).
Cf. A195168, A047218, A029919 (all have common initial terms but differ later and only remain lower resp. upper bounds).


KEYWORD

nonn


AUTHOR



STATUS

approved



