%I #19 Apr 08 2020 12:51:45
%S 1,2,4,5,7,8,10,11,13,14,15,17,18,20,21,23,24,26,27,28,30,31,33,34,36,
%T 37,39,40,41,43,44,46,47,49,50,51,53,54,56,57,59,60,62,63,64,66,67,69,
%U 70,72,73,75,76,77,79,80,82,83,85,86,88,89,90,92,93,95,96,98,99,101
%N Smallest positive integer m where 1^n+2^n+3^n+...+m^n is greater than or equal to (m+1)^n.
%C The two trivial cases of equality are n=0, m=1 and n=1, m=2, i.e. 1^0=2^0 and 1^1+2^1=3^1. The references state that there are no other equalities for m<10^2000000.
%D R. K. Guy, Unsolved Problems in Number Theory, D10.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Power.html">Power.</a>
%F Conjecture: a(n) = 1 + round(n/log(2)). Formula verified for n=1..700. - _Herbert Kociemba_, Apr 08 2020
%e a(3)=5 since 1^3+2^3+3^3+4^3<5^3 but 1^3+2^3+3^3+4^3+5^3>=6^3, i.e. since 100<125 but 225>=216.
%p A072633 := proc(n)
%p local msum,m ;
%p msum := 1;
%p m := 1 ;
%p while msum < (m+1)^n do
%p m := m+1 ;
%p msum := msum+m^n ;
%p end do:
%p return m ;
%p end proc:
%p seq(A072633(n),n=0..30) ; # _R. J. Mathar_, Feb 27 2018
%t (* Assuming sequence is increasing : *) a[0] = 1; a[n_] := a[n] = (m = a[n-1]; While[ True, m++; If[ Sum[ k^n, {k, 1, m}] >= (m+1)^n, Break[]]]; m); Table[ a[n], {n, 0, 69}] (* _Jean-François Alcover_, Oct 03 2011 *)
%Y Close to A037087 (offset).
%K nonn
%O 0,2
%A _Henry Bottomley_, Jun 28 2002