A234519 Natural numbers n sorted by decreasing values of number k(n) = sigma(n)^(1/n), where sigma(n) = A000203(n) = the sum of divisors of n.

2, 4, 3, 6, 5, 8, 7, 10, 9, 12, 14, 11, 16, 15, 18, 13, 20, 24, 17, 21, 22, 19, 28, 26, 30, 23, 25, 27, 32, 36, 34, 33, 29, 40, 31, 35, 42, 38, 39, 44, 48, 37, 45, 46, 41, 50, 54, 52, 43, 56, 60, 51, 49, 47, 55, 58, 57, 64, 66, 53, 63, 62, 72, 68, 70, 59, 65

Offset: 1

Author

Jaroslav Krizek, Jan 04 2014

Keywords

• nonn

Comments

Number k(n) = sigma(n)^(1/n) is number such that k(n)^n = sigma(n).

For number 2; k(2) = sigma(2)^(1/2) = sqrt(3) = 1,732050807568… = A002194 (maximal value of function k(n)).

The last term of this infinite sequence is number 1, k(1) = 1 (minimal value of function k(n)).

Conjecture: Every natural number n has a unique value of number k(n).

See A234521 - sequence of numbers a(n) such that a(n) > a(k) for all k < n.

Links

• Jaroslav Krizek, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A234515, A234516, A234517, A234518, A234520, A234521, A234522, A234523, A234524.

Programs

PARI

a(n)=vecsort(vector(2*n, i, sigma(i)^(1/i)), ,5)[n] \\ Michel Marcus and Ralf Stephan, Jan 14 2014