A381767 a(n) = ceiling(n^(n/(n-1))) with a(1)=1.

1, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71

Offset: 1

Author

Mike Sheppard, Mar 06 2025

Keywords

• nonn

Comments

a(n) is the smallest positive integer expressible as both the sum and product of the same set of n positive real numbers.

For a given n>1, the smallest possible sum (which is also the product) is achieved when all n real numbers are equal. Restricting to positive real numbers, let each number be c, then: n*c = c^n. Solving for c, we get: c=n^(1/(n-1)). Substituting this back into the sum (product) equation, n*c=c^n = n^(n/(n-1)). Thus, the smallest integer satisfying this condition is a(n) = ceiling(n^(n/(n-1))).

The growth of n^(n/(n-1)) means some integers are skipped in the sequence, see A382131 for those.

Examples

			For n=1 the equation 1*c=c^1 is true for all values of c. We define a(1)=1 as for any positive integer equal to or greater than 1 we have x=x (singular sum equals singular product).
For n=4, solving 4*c=c^4 with positive c yields c~1.5874, giving 4*c=c^4~6.3496. Any integer greater than 6.3496 allows for a multiset of n positive real numbers that sum to and multiply to that integer. Hence, the smallest integer satisfying the condition is a(4)=7.
A constructive procedure is to let {c1,c2} be two numbers such that c1 appears once and c2 appears (n-1) times, then c1+(n-1)*c2 = c1*c2^(n-1)=d can be solved numerically for {c1,c2} for a given {n,d}. Multiple real solutions appear.
a(4)=7
7 : {3.0373, 1.3209, 1.3209, 1.3209} or {0.788707, 2.07043, 2.07043, 2.07043}
100 : {96.9691, 1.01031, 1.01031, 1.01031} or {0.00270022, 33.3324, 33.3324, 33.3324}
		

Crossrefs

Cf. A382131 (complement).

Programs

Mathematica

Prepend[Table[Ceiling[n^(n/(n - 1))], {n, 2, 100}], 1]

PARI

a(n) = if (n==1, 1, ceil(n^(n/(n-1)))); \\ Michel Marcus, Apr 05 2025

Python

from sympy import integer_nthroot
def A381767(n): return (lambda x: x[0]+(not x[1]))(integer_nthroot(n**n,n-1)) if n>1 else 1 # Chai Wah Wu, Apr 01 2025

Formula

n^(n/(n-1)) = n + log(n) + O(log(n)^2)/n. - Yifan Xie, May 20 2025