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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.
10
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
1,1
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
PROG
(PARI) a(n)=vecsort(vector(2*n, i, sigma(i)^(1/i)), , 5)[n] \\ Michel Marcus and Ralf Stephan, Jan 14 2014
KEYWORD
nonn
AUTHOR
Jaroslav Krizek, Jan 04 2014
STATUS
approved