A332097 Maximum of s^n - Sum_{0 < x < s} x^n.

1, 1, 4, 28, 317, 4606, 84477, 1919575, 47891482, 1512466345, 48627032377, 1930020260416, 77986967769593, 3624337209819538, 178110510699972510, 9381158756438306167, 548676565488760277878, 31900481466759651567625, 2189463436999785648552851, 144075114432622269076465962

Offset: 0

Author

M. F. Hasler, May 07 2020

Keywords

• nonn

Comments

For small values of s, we have Sum_{0 < x < s} x^n ~ (s-1)^n, but for s > n/log(2) + 1.5 (cf. A332101) the difference E(s) = s^n - Sum_{0 < x < s} x^n becomes negative. Just before, the difference has its maximum: We have E(s) < E(s+1) <=> 2*s^n < (s+1)^n <=> s < 1/(2^(1/n)-1), so E takes its maximum at s = A078607(n), the least integer larger than this limiting value. This appears to be almost always equal to A332101(n) - 2.

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..367

Crossrefs

Cf. A030052 (least k such that k^n = sum of distinct n-th powers).

Cf. A078607 (s for which E(s) = a(n) <=> least k such that 2*k^n > (k+1)^n).

Cf. A332065 (all k such that k^n is a sum of distinct n-th powers).

Cf. A332101 (least k such that k^n <= sum of all smaller n-th powers).

Programs

Maple

a:= proc(n) option remember; `if`(n=0, 1, (s->
      s^n-add(x^n, x=1..s-1))(ceil(1/(2^(1/n)-1))))
    end:
seq(a(n), n=0..20);  # Alois P. Heinz, May 09 2020

Mathematica

a[0] = 1; a[n_] := (s = Ceiling[1/(2^(1/n) - 1)])^n - Sum[k^n, {k, 1, s - 1}]; Array[a, 20, 0] (* Amiram Eldar, May 09 2020 *)

PARI

{apply( A332097(n,s=1\(sqrtn(2,n-!n)-1))=(s+1)^n-sum(k=1,s,k^n), [0..20])}

Formula

a(n) = s^n - Sum_{0 < x < s} x^n for s = ceiling(1/(2^(1/n)-1)) = A078607(n).