A072633 Smallest positive integer m where 1^n+2^n+3^n+...+m^n is greater than or equal to (m+1)^n.

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, 37, 39, 40, 41, 43, 44, 46, 47, 49, 50, 51, 53, 54, 56, 57, 59, 60, 62, 63, 64, 66, 67, 69, 70, 72, 73, 75, 76, 77, 79, 80, 82, 83, 85, 86, 88, 89, 90, 92, 93, 95, 96, 98, 99, 101

Offset: 0

Author

Henry Bottomley, Jun 28 2002

Keywords

• nonn

Comments

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.

Examples

			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.
		

References

• R. K. Guy, Unsolved Problems in Number Theory, D10.

Links

• Eric Weisstein's World of Mathematics, Power.

Crossrefs

Close to A037087 (offset).

Programs

Maple

A072633 := proc(n)
    local msum,m ;
    msum := 1;
    m := 1 ;
    while msum < (m+1)^n do
        m := m+1 ;
        msum := msum+m^n ;
    end do:
    return m ;
end proc:
seq(A072633(n),n=0..30) ; # R. J. Mathar, Feb 27 2018

Mathematica

(* 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 *)

Formula

Conjecture: a(n) = 1 + round(n/log(2)). Formula verified for n=1..700. - Herbert Kociemba, Apr 08 2020