A064158 Integers m such that (x1*x2*..xk)^(x1+x2+..xk) = (x1+x2+..xk)^(x1*x2*..xk) where x1x2..xk are the digits of m in base 10.

1, 2, 3, 4, 5, 6, 7, 8, 9, 22, 112, 121, 123, 132, 211, 213, 231, 312, 321, 1124, 1142, 1214, 1241, 1412, 1421, 2114, 2141, 2411, 4112, 4121, 4211, 11125, 11133, 11152, 11215, 11222, 11251, 11313, 11331, 11512, 11521, 12115, 12122, 12151, 12212

Offset: 1

Author

Felice Russo, Sep 14 2001

Keywords

• easy
• nonn
• base
• changed

Comments

With the exception of only 112, 121, and 211, each term of this sequence satisfies (sum of digits) equals (product of digits). For 112, 121, and 211, the sum of the digits is 4, the product of the digits is 2, and the terms qualify because 2^4 equals 4^2. - Harvey P. Dale, Sep 30 2011

Examples

			22 belongs to the sequence because (2*2)^(2+2) = (2+2)^(2*2).
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..179

Crossrefs

Cf. A007953, A007954, A034710.

Programs

Mathematica

okQ[n_]:=Module[{idn=IntegerDigits[n],t,p}, t= Times@@idn; p=Total[idn];t^p==p^t]; Select[Range[12500],okQ]

Formula

Union of {112, 121, 211} and A034710. - Jason Yuen, Aug 27 2025

Extensions

More terms from Jason Earls, Dec 04 2001