A271936 Commutative powers: numbers of the form a^b = b^a.

1, 4, 16, 27, 256, 3125, 46656, 823543, 16777216, 387420489, 10000000000, 285311670611, 8916100448256, 302875106592253, 11112006825558016, 437893890380859375, 18446744073709551616, 827240261886336764177, 39346408075296537575424, 1978419655660313589123979, 104857600000000000000000000

Offset

1,2

Comments

The only integer solutions (a,b) to the equation a^b = b^a are (k,k) for k >= 0, (2,4), and (4,2). [Corrected by M. F. Hasler, Oct 15 2019]

Suppose a < b. If a^b = b^a, log(a^b) = log(b^a) so log(a)/a = log(b)/b.

Now f(x) = log(x)/x is strictly increasing at (0,e) and strictly decreasing at (e, +infinity), so if 0 < x < y < e or e < y < x then f(x) < f(y).

To have f(a) = f(b) (which means a^b = b^a), a must be in (0,e) and b in (e, +infinity). Let g(a) be the restriction of f to (0,e) and h(b) to (e, +infinity). Since g and h are monotonic, there is a natural bijection between a and b such that g(a) = h(b).

In (0,e) there are two integers: 1 and 2.

1 won't work since there are no solutions 1^b=b^1 for b>e.

2 will because 2^4 = 4^2 and since we have a bijection this is the only solution.

This is the range of A000312(n) = n^n together with { 16 }. - M. F. Hasler, Oct 15 2019

Links

Table of n, a(n) for n=1..21.

Formula

a(n) = A000302(n-1) = 4^(n-1) for 1 <= n <= 3;

a(n) = A000312(n-1) = (n-1)^(n-1) for all n >= 4.

Example

4 is in the list because 2^2 = 2^2;
16 is in the list because 2^4 = 4^2. This is the only term not of the form n^n.

Mathematica

Union[Power @@@ Select[Tuples[Range@ 20, 2], Power @@ # == Power @@ Reverse@ # &]] (* Michael De Vlieger, Apr 17 2016 *)

Prog

(PARI) a(n)=if(n>3, n-1, 4)^(n-1) \\ M. F. Hasler, Oct 15 2019

Crossrefs

Cf. A000312.

Sequence in context: A067688 A097374 A257309 * A046358 A046366 A227609

Adjacent sequences: A271933 A271934 A271935 * A271937 A271938 A271939

Keyword

nonn,easy

Author

Natan Arie Consigli, Apr 16 2016

Extensions

Edited by M. F. Hasler, Oct 15 2019

Status

approved